Prove that 16 divides n4 + 4n2 + 11, if n is an odd integer.
Given here, n is an odd integer for some k Z
where Z is the set of all integers.
Since, we know that every odd integer is of the form 4k + 1 and 4k – 1.
Consider two cases:
Case 1: For n = 4k + 1
)
Therefore, it is divisible by 16.
Case 2: For n = 4k – 1
)
Therefore, it is divisible by 16.
Thus, n4 + 4n2 + 11 is divisible by 16.
Prove that if n is a positive even integer, then 24 divides n(n + 1)(n + 2).
Given here, n is a positive even integer.
We know that any positive even integer can be expressed as n = 2k
Therefore, n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)……eq (1)
Now, considering the following cases,
Case 1: k = 1
n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)
= 2 × 1(2 × 1 + 1)(2 × 1 + 2)
= 2 × (3) × (4)
= 24
Hence, n(n + 1)(n + 2) is divisible by 24. (From (1))
Case 2: k = 2
n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)
= 2 × 2(2 × 2 + 1)(2 × 2 + 2)
= 4 × (5) × (6)
= 24 × 5
Hence, n(n + 1)(n + 2) is divisible by 24. (from (1))
Case 3: k3
Here, k and k + 1 being even integers, one of them will always be odd and the other will be even.
So, 2k(2k + 1)(2k + 2) will always be divisible by 2.
Also as k≥ 3 is a positive integer, so for some l∈N,
k = 3a or k = 3a + 1 or k = 3a + 2
For k = 3a
k(k + 1)(2k + 1) = 3a(3a + 1)(2(3a) + 1)
= 3(a(3a + 1)(6a + 1)
Therefore, it is divisible by 3.
For k = 3a + 1
k(k + 1)(2k + 1) = (3a + 1)(3a + 1 + 1)(2(3a + 1) + 1)
= (3a + 1)(3a + 2)(6a + 3)
= 3(3a + 1)(3a + 2)(2a + 1)
Therefore, it is divisible by 3.
For k = 3a + 2
k(k + 1)(2k + 1) = (3a + 2)(3a + 2 + 1)(2(3a + 2) + 1)
= (3a + 2)(3a + 3)(6a + 5)
= 3(3a + 2)(a + 1)(6a + 5)
Therefore, it is divisible by 3.
We can see that in any case k(k + 1)(2k + 1) is divisible by 2 and 3.
Also, we know that 2 and 3 are mutually prime numbers.
Therefore the expression must be divisible by 2 × 3 = 6
So, k(k + 1)(2k + 1) is divisible by 24. (From (1))
Prove that if either of 2a + 3b and 9a + 5b is divisible by 17, so is the other. a, b E N (Hint: 4(2a + 3b) + 9a + 5b = 17a + 17b)
Let 2a + 3b be divisible by 17.
Therefore, for some integer k, 2a + 3b = 17k….. eq (1)
Now,
9a + 5b = 17a + 17b – 4(2a + 3b)
= 17(a + b) – 4(17k) (from eq (1))
= 17(a + b – 4k)
Therefore, we can say that 9a + 5b is divisible by 17.
Prove that every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k E N V {0}.
We know that from Euclid’s division lemma, for every natural number a if we take b = 5 then unique non-negative integers k and m can be obtained so that a = 5k + m
Where 0 ≤ m ≤ 5
Therefore, m = 0, 1, 2, 3, 4 and 5
So, we can say that a = 5k or a = 5k + 1 or a = 5k + 2 or a = 5k + 3 or a = 5k + 4 where k E N V {0}
Now, a = 5k + 3 = 5k + 5 – 2 = 5(k + 1) – 2
= 5k1 – 2 where (k1 = k + 1)
Therefore, a = 5k + 3 can be written as a = 5k – 2
Now, a = 5k + 4 = 5k + 5 – 1 = 5(k + 1) – 1
= 5k1 – 1 where (k1 = k + 1)
Therefore, a = 5k + 4 can be written as a = 5k – 1
So, every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k E N V {0}.
Prove that if 6 has no common factor with n, n2 — 1 is divisible by 6.
Given here, there is no common factor between 6 and n.
Therefore, 6 and n are two distinct natural numbers.
We know that 6 has 2 and 3 as prime factors.
n can be written as n = 2k + 1 for all
Therefore, we can see that is divisible by 2.
Similarly, n can be written as n = 3k + 1 for all
Therefore, we can see that is divisible by 3.
Similarly, n can also be written as n = 3k–1 for all
Therefore, we can see that is divisible by 3.
So, n2 — 1 is divisible by 2 and 3 both.
Since, 2 and 3 are prime numbers.
Therefore, n2 — 1 is divisible by 2 × 3 = 6.
Prove that product of four consecutive positive integers is divisible by 24.
Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).
Product = n(n + 1)(n + 2)(n + 3)
We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.
For, n = 3k
Product = n(n + 1)(n + 2)(n + 3)
= 3k(3k + 1)(3k + 2)(3k + 3)
= 33k(3k + 1)(3k + 2)(k + 1)
Therefore it is divisible by 3.
For, n = 3k + 1
Product = n(n + 1)(n + 2)(n + 3)
= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)
= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)
Therefore it is divisible by 3.
For, n = 3k + 2
Product = n(n + 1)(n + 2)(n + 3)
= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)
= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)
Therefore it is divisible by 3.
n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2
For n = 4p
Product = n(n + 1)(n + 2)(n + 3)
= 4p(4p + 1)(4p + 2)(4p + 3)
= 24p(4p + 1)(2p + 1)(4p + 3)
= 8p(4p + 1)(2p + 1)(4p + 3)
Therefore it is divisible by 8.
For n = 4p + 1
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)
= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)
= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)
Therefore it is divisible by 8.
For n = 4p + 2
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)
= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)
= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)
Therefore it is divisible by 8.
For n = 4p + 2
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)
= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)
= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)
Therefore it is divisible by 8.
For n = 4p + 2
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)
= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)
= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)
Therefore it is divisible by 8.
For n = 4p + 3
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)
= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)
= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)
Therefore, it is divisible by 8.
Since it is divisible by both 3 and 8 and both are mutually prime numbers .
Therefore, it will be divisible by 38 = 24
So, product of four consecutive positive integers is divisible by 24.
Find g. c. d
144, 233
144, 233
Here, 233 > 144
233 = 144 × 1 + 89
144 = 89 × 1 + 55
89 = 55 × 1 + 34
55 = 34 × 1 + 21
34 = 21 × 1 + 13
21 = 13 × 1 + 8
13 = 8 × 1 + 5
8 = 5 × 1 + 3
5 = 3 × 1 + 2
3 = 2 × 1 + 1
2 = 1 × 2 + 0
The last non- zero remainder is 1.
Therefore, g. c. d (144, 233) = 1
Find g. c. d
765, 65
765, 65
Here 765 > 65
765 = 65 × 11 + 50
65 = 50 × 1 + 15
50 = 15 × 3 + 5
15 = 5 × 3 + 0
The last non- zero remainder is 5.
Therefore, g. c. d (765, 65) = 5
Find g. c. d
10211, 2517
10211, 2517
Here, 10211 > 2517
10211 = 3517 × 4 + 143
2517 = 143 × 17 + 86
143 = 86 × 1 + 57
86 = 57 × 1 + 29
57 = 29 × 1 + 28
29 = 28 × 1 + 1
28 = 1 × 28 + 0
The last non- zero remainder is 1.
Therefore, g. c. d (10211, 2517) = 1
Find g. c. d. of 736 and 85 by using Euclid's algorithm.
Here, 736 > 85
736 = 85 × 8 + 56
85 = 56 × 1 + 29
56 = 29 × 1 + 27
29 = 27 × 1 + 2
27 = 2 × 13 + 1
2 = 1 × 2 + 0
The last non- zero remainder is 1.
Therefore, g. c. d (736, 85) = 1
Prove g. c. d (a - b, a + b) = 1 or 2, if g. c. d. (a, b) = 1
Given g.c.d. (a,b) = 1
From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b
Let g. c. d (a - b, a + b) = k
Therefore we can say that k is a factor of both (a–b) and (a + b).
We can write a–b = rk for some r N
And also a + b = sk for some sN
Now, (a + b) + (a–b) = rk + sk
a + a + b–b = k(r + s)
2a = k(r + s)………..eq(1)
(a + b) – (a – b) = rk – sk
a + b–a + b = k(r–s)
2b = k(r–s)………..eq(2)
Also, g. c. d (a, b) = 1
Therefore, 2 × g. c. d (a, b) = 2 × 1
g. c. d (2a, 2b) = 2
g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))
k × g. c. d(r + s, r – s) = 2
= 2 × 1
So, k × g. c. d(r + s, r – s) = 2
k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)
By comparing we get k = 2
We know that 1 and 2 are co–prime numbers.
Similarly, we get k = 1
So, g. c. d (a—b, a + b) = k = 2
or g. c. d (a—b, a + b) = k = 1
g. c. d (a—b, a + b) = 1 or 2
Using the fact that g. c. d (a, b) l.c.m.. (a, b) = ab, find l.c.m.. (115, 25)
Here, 115 > 25
115 = 25 × 4 + 15
25 = 15 × 1 + 10
15 = 10 × 1 + 5
10 = 5 × 2 + 0
The last non- zero remainder is 5.
Therefore, g. c. d(115, 25) = 5
Now, by Euclid’s Algorithm
g. c. d(a, b) × l.c.m.(a, b) = ab
Here, a = 115, b = 25
g. c. d(115, 25) × l.c.m.(115, 25) = 115 × 25
⇒ 5 × l.c.m.(115, 25) = 115 × 25
⇒ l.c.m.(115, 25) =
⇒ l.c.m.(115, 25) = 115 × 5
Therefore, l.c.m.(115, 25) = 575
Express as a product of primes:
7007
7007 = 7 × 1001
= 7 × 7 × 143
= 7 × 7 × 11 × 13
= 72 × 11 × 13
Therefore, 7007 can be expressed as a product of prime numbers 7, 11 and 13.
Express as a product of primes:
7500
7500 = 2 × 3750
= 2 × 2 × 1875
= 2 × 2 × 3 × 625
= 2 × 2 × 3 × 5 × 125
= 2 × 2 × 3 × 5 × 5 × 25
= 2 × 2 × 3 × 5 × 5 × 5 × 5
= 22 × 3 × 54
Therefore, 7500 can be expressed as a product of prime numbers 2, 3 and 5.
Express as a product of primes:
10101
10101 = 3 × 3367
= 3 × 7 × 481
= 3 × 7 × 13 × 37
Therefore, 10101 can be expressed as a product of prime numbers 3, 7, 13 and 37
Express as a product of primes:
15422
15422 = 2 × 7711
= 2 × 11 × 701
Therefore, 15422 can be expressed as a product of prime numbers 2, 11 and 701
Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:
250 and 336
250 = 2 × 125
= 2 × 5 × 25
= 2 × 5 × 5 × 5 = 2 × 55
336 = 2 × 168
= 2 × 2 × 84
= 2 × 2 × 2 × 42
= 2 × 2 × 2 × 2 × 21
= 2 × 2 × 2 × 2 × 3 × 7
= 24 × 3 × 7
Therefore, g. c. d(250, 336) = 2
l.c.m.(250, 336) = 24 × 3 × 7 × 53
= 16 × 21 × 125
= 336 × 125 = 42000
Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:
4000 and 25
4000 = 2 × 2000
= 2 × 2 × 1000
= 2 × 2 × 2 × 500
= 2 × 2 × 2 × 2 × 250
= 2 × 2 × 2 × 2 × 2 × 125
= 2 × 2 × 2 × 2 × 2 × 5 × 25
= 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
= 25 × 53
25 = 5 × 5
g. c. d(4000, 25) = 5 × 5 = 25
l.c.m.(4000, 25) = 25 × 53 = 4000
Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:
225 and 145
225 = 5 × 45 = 5 × 5 × 9 = 5 × 5 × 3 × 3
145 = 5 × 29
g. c. d(225, 145) = 5
l.c.m.(225, 145) = 52 × 32 × 29
= 25 × 9 × 29= 6525
Find g. c. d. and l.c.m.. using the fundamental theorem of arithmetic:
175 and 1001
175 = 5 × 35 = 5 × 5 × 7
1001 = 7 × 143 = 7 × 11 × 13
g. c. d(175, 1001) = 7
l.c.m.(175, 1001) = 52 × 7 × 11 × 13
= 25025
Find g. c. d. and l.c.m.. :
15, 21, 35
15 = 1 × 3 × 5
21 = 1 × 3 × 7
35 = 1 × 5 × 7
Therefore, g. c. d(15, 21, 35) = 1
l.c.m.(15, 21, 35) = 3 × 5 × 7 = 15 × 7 = 105
Find g. c. d. and l.c.m.. :
40, 60, 80
40 = 2 × 20 = 2 × 2 × 10
= 2 × 2 × 2 × 5
60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 5 × 3
80 = 2 × 40 = 2 × 2 × 20 = 2 × 2 × 2 × 10
= 2 × 2 × 2 × 2 × 5
Therefore, g. c. d(40, 60, 80) = 2 × 2 × 5 = 20
l.c.m.(40, 60, 80) = 24 × 3 × 5 = 16 × 15 = 240
Find g. c. d. and l.c.m.. :
49, 42, 91
49 = 7 × 7
42 = 2 × 21 = 2 × 3 × 7
91 = 7 × 13
Therefore, g. c. d(49, 42, 91) = 7
l.c.m.(49, 42, 91) = 72 × 2 × 3 × 13
= 49 × 78 = 3822
Prove that following numbers are irrational:
Let be a rational number.
Therefore,, where g. c. d(a, b) = 1 and a, b N.
Now, squaring both sides
……. eq (1)
5 | a2
5 | a
Now, let a = 5k, k N
So, b = 5k
5 | b
It is a contradiction as g. c. d (a, b) = 1
So, our assumption is wrong.
Therefore, is irrational.
Prove that following numbers are irrational:
Let be a rational number.
Therefore , where g. c. d(a, b) = 1 and a, b N.
Now, squaring both sides
……. eq (1)
15 | a2
15 | a
Now, let a = 15k, k N
So, b = 15k
15 | b
It is a contradiction as g. c. d(a, b) = 1
So, our assumption is wrong.
Therefore, is irrational.
Prove that following numbers are irrational:
.
Let k = is a rational number
Therefore, = 1 which is also a rational number
But, is irrational and we know that subtraction of a rational and irrational is always irrational.
So, our assumption is wrong.
is an irrational number.
Prove that following numbers are irrational:
Let k = is a rational number
Therefore,
is rational
is also rational
But, we know that is irrational.
So, our assumption is wrong.
is an irrational number.
Prove that following numbers are irrational:
Let k = is a rational number
Therefore, = is also a rational number
But, is irrational.
So, our assumption is wrong.
is an irrational number
Find l.c.m.. (105, 91) using g. c. d. (a, b) (a, b) = ab
Here, 105 > 91
105 = 91 × 1 + 14
91 = 14 × 6 + 7
14 = 7 × 2 + 0
Last non-negative remainder is 7.
Therefore, g. c. d(105, 91) = 7
Now, we know the relation
g. c. d(a, b) × l.c.m.(a, b) = ab
putting a = 105 and b = 91
g. c. d(105, 91) × l. c. d(105, 91) = 105 × 91
7 × l. c. d(105, 91) = 105 × 91
l. c. d(105, 91) =
l. c. d(105, 91) = 105 × 13 = 1365
Using (√7 + √3)(√7 — √3) = 4 and the fact that (√7 + √3) is irrational prove that √7 — √3 is irrational.
Given ( + )( — ) = 4 and ( + ) is irrational.
(multiplying numerator and denominator by )
=
(we know that (a – b) × (a + b) = (a2 – b2))
= =
Here, it is given that is irrational and we know that 4 is rational.
Therefore,
We know that rational divided by irrational is always irrational.
So, is irrational.
Two buses start from the same spot for the same circular root. One is a BRTS bus returning in 35 minutes. The other is a regular express bus taking 42 minutes to return. After how many minutes will they meet again at the same initial spot?
Given that the BRTS bus returns in 35 minutes and the express bus returns in 42 minutes.
So, we can say that the BRTS returns in the time intervals which are multiples of 35 and the express bus returns in the time intervals which are multiples of 42.
Therefore, both the buses meet at the time interval of l.c.m. of 35 and 42.
35 = 5 × 7
42 = 2 × 3 × 7
l.c.m.(35, 42) = 5 × 2 × 3 × 7 = 210 minutes
So, both the buses will meet after 210 minutes at the initial spot.
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 625 = 5 × 5 × 5 × 5 = 54 × 1 = 54 × 20
i.e. it is in the power of 5
So, the given rational number is terminating in nature.
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 3125 = 5 × 5 × 5 × 5 × 5 = 55
i.e. it is in the power of 5
So, the given rational number is terminating in nature.
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
So, the given rational number is terminating in nature.
Prime factorization of 6250 = 5 × 5 × 5 × 5 × 5 × 2 = 55 × 21
i.e. it is in the power of 5 & 2
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 15625 = 5 × 5 × 5 × 5 × 5 × 5 × 5 = 57
i.e. it is in the power of 5
So, the given rational number is terminating in nature
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 500 = 5 × 5 × 5 × 2 × 2 =
i.e. it is in the power of 5 & 2
So, the given rational number is terminating in nature
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 =
i.e. it is in the power of 5 & 2
So, the given rational number is terminating in nature
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
can be written as
Prime factorization of 26 = 2 × 13
Here, denominator has got a factor other than 2 &5
So, it is non-terminating in nature.
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
can be written as
As, its denominator is itself 5. So, it is terminating in nature .
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 343 = 7 × 7 × 7
i.e. it is not in the power of 5 & 2
So, the given rational number is non- terminating in nature.
State whether following rational number have terminating decimal expansion or not and if it has terminating decimal expansion, find it:
Convert the given rational number into lowest divisible terms.
If the denominator is in the power of 2 OR/AND 5, then the given number is terminating else non- terminating.
Prime factorization of 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27
i.e. it is in the power of 2.
So, the given rational number is terminating in nature .
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
0. 01001000100001….
The given decimal number is non-recurring and non-terminating Therefore, it’s an irrational number.
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
Let × = 3. 456789123456…………eq(1)
As there are 9 repeating digits. So we multiply by 109
109 × = 3456789123. 45678912345678…. …..eq(2)
Subtracting (1)from(2):
99999999 × = 3456789120
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
5. 123456789
This is a terminating number so its rational for
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
Let 10 × = 23. 12121212…. ….eq(1)
As there are 2 repeating digits. So we multiply by 102
1000 × = 2312. 121212……….eq(2)
Subtracting (1) from (2):
990 × = 2289
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
Let × = 0. 142857142857142…. …eq(1)
As there are 6 repeating digits. So we multiply by 106
106 × = 142857. 14285714…. …..eq(2)
Subtracting (1) from (2):
99999 × = 142857
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
Let × = 0. 9999…. ….eq(1)
As there are repeating digits.
So we multiply by 10
10 × = 9. 9999…. ……..eq(2)
Subtracting (1) from (2):
9 × = 9
× = 1
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
5. 781
As the given number is non-terminating and non-recurring so its rational number can be shown as,
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
2. 312
As the given number is non-terminating and non-recurring so its rational number can be shown as,
Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational. express them in form p/q. Comment on factors of q:
0. 12345
As the given number is non-terminating and non-recurring so its rational number can be shown as,
Find the square roots of following surds:
Let =
Now, analysing × and y such that × + y = 5 and × y = 6
Therefore, it can be found that × = 3 and y = 2
So, =
Find the square roots of following surds:
Let =
Now, analysing × and y such that × + y = 9 and × y = 14
Therefore, it can be found that × = 7 and y = 2
So, =
Find the square roots of following surds:
Let =
Now, analysing × and y such that × + y = 2 and × y =
Therefore, it can be found that × = 3 and y = 2
So, =
Find the square roots of following surds:
=
=
=
Find the square roots of following surds:
=
=
=
=
=
Find the square roots of following surds:
=
=
=
=
=
Find the square roots of following surds:
=
=
=
Simplify:
=
=
=
= ………….. eq (1)
=
=
=
= ………….. eq (2)
=
=
=
= ………………eq (3)
Now, subtracting eq (3) from the addition of eq (1) and eq (2).
+ =
+
Now, rationalising the terms
+
= +
= +
= +
= +
= +
=
= =
Find g. c. d and l.c.m.
25, 35
25 = 5 × 5
35 = 5 7
Therefore, g. c. d(25, 35) = 5
And l.c.m. (25, 35) = 5 × 7 × 5 = 175
Find g. c. d and l.c.m.
105, 125
105 = 3 × 35 = 3 × 5 × 7
125 = 5 × 25 = 5 × 5 × 5
Therefore, g. c. d(105, 125) = 5
And l.c.m.(105, 125) = 3 × 5 × 5 × 5 × 7 = 2675
Find g. c. d and l.c.m.
220, 132
220 = 2 × 110 = 2 × 2 × 55 = 2 × 2 × 5 × 11
132 = 2 × 66 = 2 × 2 × 33 = 2 × 2 × 3 × 11
Therefore, g. c. d(220, 132) = 2 × 2 × 11 = 44
And l.c.m.(220, 132) = 2 × 2 × 3 × 11 × 5 = 660
Find g. c. d and l.c.m.
3125, 625
3125 = 5 × 625 = 5 × 5 × 125 = 5 × 5 × 5 × 5 × 5
625 = 5 × 125 = 5 × 5 × 5 × 5
Therefore, g. c. d(3125, 625) = 5 × 5 × 5 × 5 = 625
And l.c.m.(3125, 625) = 5 × 5 × 5 × 5 × 5 = 3125
Find g. c. d and l.c.m.
15625, 35
15625 = 5 × 5 × 5 × 5 × 5 × 5
35 = 5 × 7
Therefore, g. c. d(15625, 625) = 5
And l.c.m.(15625, 625) = 5 × 5 × 5 × 5 × 5 × 5 × 7 = 109375
Find g. c. d and l.c.m.
15, 25, 35
15 = 5 × 3
25 = 5 × 5
35 = 5 × 7
Therefore, g. c. d(15, 25, 35) = 5
And l.c.m.(15, 25, 35) = 5 × 5 × 3 × 7 = 525
Find g. c. d and l.c.m.
18, 12, 16
18 = 3 × 3 × 2
12 = 3 × 2 × 2
16 = 2 × 2 × 2 × 2
Therefore, g. c. d(18, 12, 16) = 2
And l.c.m.(18, 12, 16) = 2 × 2 × 2 × 2 × 3 × 3 = 144
Find g. c. d and l.c.m.
16, 24, 36
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
36 = 2 × 3 × 2 × 3
Therefore, g. c. d(16, 24, 36) = 2 × 2 = 4
And l.c.m.(16, 24, 36) = 2 × 2 × 2 × 2 × 3 × 3 = 144
Find g. c. d and l.c.m.
35, 28, 63
35 = 5 × 7
28 = 2 × 2 × 7
63 = 3 × 3 × 7
Therefore, g. c. d(35, 28, 63) = 7
And l.c.m.(35, 28, 63) = 3 × 3 × 2 × 2 × 7 × 5 = 1260
Find g. c. d and l.c.m.
112, 128, 144
For L.C.M or G.C.D we first need to find the prime factors of the numbers
112 = 2 × 2 × 2 × 2 × 7
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
144 = 2 × 2 × 2 × 2 × 3 × 3
Therefore, g. c. d(112, 128, 144) = 2 × 2 × 2 × 2 = 16
L.C.M of a group of numbers is the product of all the factors of three numbers without replicating the number more than one which is common to all numbersProve following numbers are irrational.
Let k = is a rational number
Therefore,
Squaring both sides we have,
is rational
is also rational
But, we know that is irrational.
So, our assumption is wrong.
is an irrational number.
Prove following numbers are irrational.
Let k = is a rational number
Therefore, = is also a rational number
But, is irrational.
So, our assumption is wrong.
is an irrational number
Prove following numbers are irrational.
Let is a rational number
Therefore, k =
Squaring both sides we have,
is rational
is also rational
But, we know that is irrational.
So, our assumption is wrong.
is an irrational number.
So, is also an irrational number.
Prove following numbers are irrational.
Let k = is a rational number
Therefore,
Squaring both sides we have,
is rational
is also rational
But, we know that is irrational.
So, our assumption is wrong.
is an irrational number.
Prove following numbers are irrational.
Let k = is a rational number
Therefore, is a rational number
But, we know that is irrational.
So, our assumption is wrong.
is an irrational number.
Prove following numbers are irrational.
Let k = is a rational number
Therefore,
Squaring both sides we have,
is rational
is also rational
But, we know that is irrational.
So, our assumption is wrong.
is an irrational number.
Prove following numbers are irrational.
Let k =
Squaring both sides we have,
is rational
is also rational
But, we know that is irrational.
So, our assumption is wrong.
Therefore, is an irrational number.
Prove following numbers are irrational.
Let k = is a rational number
Therefore, = is also a rational number
But, is irrational.
So, our assumption is wrong.
is an irrational number.
So, is an irrational number.
Prove following numbers are irrational.
Let k = is a rational number
Therefore, = is also a rational number
But, is irrational.
So, our assumption is wrong.
is an irrational number.
So, is also an irrational number.
Prove following numbers are irrational.
We know that 37 is a prime number.
Therefore is an irrational number as it has only two factors 37 and 1.
Which of the following numbers have terminating decimal expansion and why?
Here, the denominator is 125.
125 = 5 × 5 × 5
We can see that that 125 contains 5 as its prime factors.
Also, 211 is a prime number.
So, g. c. d(211, 125) = 1
Therefore,
has a terminating decimal expansion.
Which of the following numbers have terminating decimal expansion and why?
Here, the denominator is 625.
625 = 5 × 5 × 5 × 5 × 20
We can see that that 625 contains 5 as its prime factors.
Also, 156 = 2 × 2 × 3 × 13
So, g. c. d(156, 625) = 1
Therefore,
has a terminating decimal expansion.
Which of the following numbers have terminating decimal expansion and why?
Here, the denominator is 35.
35 = 5 × 7 × 20
We can see that that 35 contains 7 as its prime factors other than 5 and 2.
Also, 337 is a prime number.
So, g. c. d(337, 35) = 1
Therefore,
does not have a terminating decimal expansion.
Which of the following numbers have terminating decimal expansion and why?
Here, the denominator is 49.
49 = 7 × 7 × 20 × 50
We can see that that 49 contains 7 as its prime factors other than 2 and 5.
Also, 132 = 2 × 2 × 3 × 11
So, g. c. d(132, 49) = 1
Therefore,
does not have a terminating decimal expansion
Which of the following numbers have terminating decimal expansion and why?
Here, the denominator is 16.
16 = 2 × 2 × 2 × 2 × 50
We can see that that 16 contains 2 as its prime factors.
Also, 235 = 5 × 47
So, g. c. d(235, 16) = 1
Therefore,
has a terminating decimal expansion.
Find the square root of the following in the form of a binomial surd :
Let =
Now, analysing x and y such that x + y = 12 and xy = 35
Therefore, it can be found that x = 7 and y = 5
So, =
Find the square root of the following in the form of a binomial surd :
Let =
Now, analysing x and y such that x + y = 8 and xy = 7
Therefore, it can be found that x = 7 and y = 1
So, =
Find the square root of the following in the form of a binomial surd :
Let =
Now, analysing x and y such that x + y = 6 and xy = 5
Therefore, it can be found that x = 5 and y = 1
So, =
Find the square root of the following in the form of a binomial surd :
Let =
= =
Now, analysing x and y such that x + y = 14 and xy = 45
Therefore, it can be found that x = 9 and y = 5
So, =
Find the square root of the following in the form of a binomial surd :
=
=
=
Simplify :
+
Now, rationalising the terms
= +
= +
= +
=
=
Simplify :
=
=
=
=
=
=
=
=
=
Find the largest number dividing 230 and 142 and leaving remainders 5 and 7 respectively.
Let k be the largest number dividing 230 and 142 leaving remainders 5 and 7 respectively.
Therefore, we can write 230 = ka + 5 and 142 = kb + 7
Where, a, b
230 = ka + 5 142 = kb + 7
⇒ ka = 230 – 5 ⇒ kb = 142 – 7
ka = 225 kb = 135
But, k is the largest number.
So, k becomes the largest divisor, g. c. d of 225 and 135
225 = 5 × 5 × 3 × 3
135 = 5 × 3 × 3 × 3
Therefore, g. c. d(225, 135) = k
g. c. d(225, 135) = 5 × 3 × 3 = 45
k = 45
Therefore, the largest number dividing 230 and 142 leaving remainders 5 and 7 respectively is 45.
Find the largest number dividing 110, 62, 92 and leaving remainders 5, 6 and 1 respectively.
Let k be the largest number dividing 110, 62 and 92 leaving remainders 5, 6 and 1 respectively.
Therefore, we can write 110 = ka + 5, 62 = kb + 6 and 92 = kc + 1
Where, a, b and c
110 = ka + 5 62 = kb + 6 92 = kc + 1
⇒ ka = 110 – 5 ⇒ kb = 62 – 6 ⇒ kc = 92 – 1
ka = 105 kb = 56 kc = 91
But, k is the largest number.
So, k becomes the largest divisor, g. c. d of 105, 56 and 91
105 = 5 × 3 × 7
56 = 7 × 2 × 2 × 2
91 = 7 × 13
Therefore, g. c. d(105, 56, 91) = k
g. c. d(105, 56, 91) = 7
k = 7
Therefore, the largest number dividing 110, 62 and 92 leaving remainders 5, 6 and 1 respectively is 7.
The length and the breadth and the height of a room are 735 cm, 625 cm and 415 cm. Find the length of the largest scale measuring instrument which can measure all the three dimensions.
The length of the largest scale measuring instrument which can measure all the three dimensions is the g. c. d of their measures.
735 = 5 × 3 × 7 × 7
635 = 5 × 5 × 5 × 5
415 = 5 × 83
Therefore, g. c. d(735, 625, 415) = 5
So, the length of the largest scale is 5 cm.
A milk man has 150 litres of milk of higher fat and 240 litres of milk of lower fat. He wants to pack the milk in tins of equal capacity. What should be the capacity of each tin?
For packing 150 litres of milk of higher fat and 240 litres of milk of lower fat in tins of equal capacity, we need to find the g. c. d of 150 and 240.
150 = 5 × 3 × 5 × 2
240 = 3 × 2 × 2 × 2 × 2 × 5
Therefore, g. c. d(150, 240) = 5 × 3 × 2 = 30
So, the capacity of the tin should be 30 litres.
Find the smallest number which decreased by 15 is a multiple of 125 and 225.
The smallest number which is a multiple of 125 and 225 is the l.c.m. of them.
125 = 5 × 5 × 5
225 = 5 × 5 × 3 × 3
Therefore, l.c.m.(125, 225) = 5 × 5 × 5 × 3 × 3 = 1125
Now, we have to find the smallest number which is decreased by 15.
Therefore, the required number = 1125 + 15 = 1140
Find the smallest number of six digits divisible by 18, 24 and 30
To find the smallest number of six digits divisible by 18, 24 and 30 we find the l.c.m. of them.
18 = 3 × 3 × 2
24 = 2 × 2 × 2 × 3
30 = 3 × 2 × 5
Therefore, l.c.m.(18, 24, 30) = 2 × 2 × 2 × 3 × 3 × 5 = 360
We know that the smallest six digit number is 100000.
Dividing 100000 by 18 = = 277. 77
So, the integer bigger and nearest to 277. 77 is 278.
Therefore, required integer = 278 × 360 = 100080
So, the smallest number of six digits divisible by 18, 24 and 30
is 100080.
Prove if 3 | (a2 + b2) then 3 |a and 3| b, a N, bN.
Let us suppose 3 is a factor of a and b.
Therefore, we can write a = 3k and b = 3m
Now, a2 + b2 = (3k)2 + (3m)2
= 9k2 + 9m2
= 3(3k2 + 3m2)
Therefore it is divisible by 3.
So, 3 is a factor of a2 + b2
Similarly, the reverse condition is also true.
Prove n4 + 4 is a composite number for n > 1
(as a2–b2 = (a + b)(a–b))
….. eq (1)
Now, n > 1
n – 1 > 0
Also,
Therefore, and are distinct positive integers.
So, we can say that
n–1 > 0
Similarly, n + 1 > 0
Therefore, are also distinct.
Thus, from eq(1) n4 + 4 has two distinct factors and 1 as a factor.
So, n4 + 4 is a composite number for n > 1.
In a morning walk a man, a woman and a child step off together. Their steps measure 90 cm, 80 cm and 60 cm. What is the minimum distance each should walk to cover the distance in complete steps?
Steps of the man measure = 90 cm
Steps of the woman measures = 80 cm
Steps of the child measures = 60 cm
To find the minimum distance each man, woman and child should walk we have to find the l.c.m. of their steps.
90 = 2 × 5 × 3 × 3
80 = 2 × 2 × 2 × 2 × 5
60 = 2 × 5 × 2 × 3
Therefore, l.c.m.(90, 80, 60) = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720
So, the minimum distance each should walk to cover the distance in complete steps is 720 cm.
Find the number nearest to 24001 and between 24001 and 25000 divisible by 16, 24, 40
To find the required number we must find the l.c.m. of 16, 24 and 40.
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
40 = 2 × 2 × 2 × 5
l.c.m.(16, 24, 40) = 2 × 2 × 2 × 2 × 3 × 5 = 240
To find the number between 24001 and 25000,
Divide 24001 by 240
= 100. 004
So, the integer nearer to and greater than it is 101.
Hence, a number divisible by 240 and between 24001 and 25000 = 101 × 240 = 24240
Product of any four consecutive positive integers is divisible by ……
A. 16
B. 48
C. 24
D. 32
Let the four consecutive integers be n, (n + 1), (n + 2) and (n + 3).
Product = n(n + 1)(n + 2)(n + 3)
We know that every integer can be written in the form of 3k, 3k + 1 and 3k + 2.
For, n = 3k
Product = n(n + 1)(n + 2)(n + 3)
= 3k(3k + 1)(3k + 2)(3k + 3)
= 33k(3k + 1)(3k + 2)(k + 1)
Therefore it is divisible by 3.
For, n = 3k + 1
Product = n(n + 1)(n + 2)(n + 3)
= 3k + 1(3k + 1 + 1)(3k + 1 + 2)(3k + 1 + 3)
= 3(3k + 1)(3k + 2)(k + 1)(3k + 4)
Therefore it is divisible by 3.
For, n = 3k + 2
Product = n(n + 1)(n + 2)(n + 3)
= 3k + 2(3k + 2 + 1)(3k + 2 + 2)(3k + 2 + 3)
= 3(3k + 2)(k + 1)(3k + 4)(3k + 5)
Therefore it is divisible by 3.
n can also be expressed as 4p, 4p + 1, 4p + 2 and 4p + 2
For n = 4p
Product = n(n + 1)(n + 2)(n + 3)
= 4p(4p + 1)(4p + 2)(4p + 3)
= 24p(4p + 1)(2p + 1)(4p + 3)
= 8p(4p + 1)(2p + 1)(4p + 3)
Therefore it is divisible by 8.
For n = 4p + 1
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 1)(4p + 1 + 1)(4p + 1 + 2)(4p + 1 + 3)
= 24(4p + 1)(2p + 1)(4p + 3)(p + 1)
= 8p(4p + 1)(2p + 1)(4p + 3)(p + 1)
Therefore it is divisible by 8.
For n = 4p + 2
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)
= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)
= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)
Therefore it is divisible by 8.
For n = 4p + 2
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)
= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)
= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)
Therefore it is divisible by 8.
For n = 4p + 2
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 2)(4p + 2 + 1)(4p + 2 + 2)(4p + 2 + 3)
= 24(2p + 1)(4p + 3)(p + 1)(4p + 5)
= 8p(2p + 1)(4p + 3)(p + 1)(4p + 5)
Therefore it is divisible by 8.
For n = 4p + 3
Product = n(n + 1)(n + 2)(n + 3)
= (4p + 3)(4p + 3 + 1)(4p + 3 + 2)(4p + 3 + 3)
= 24(4p + 3)(p + 1)(4p + 5)(2p + 3)
= 8p(4p + 3)(p + 1)(4p + 5)(2p + 3)
Therefore it is divisible by 8.
Since it is divisible by both 3 and 8 and both are mutually prime numbers .
Therefore, it will be divisible by 38 = 24
So, product of four consecutive positive integers is divisible by 24.
√4 + 3 is……..
A. irrational
B. rational but not integer.
C. non-recurring decimal
D. integer
which is an integer.
If g. c. d of two numbers is 8 and their product is 384, then their l.c.m.. is
A. 24
B. 16
C. 48
D. 32
l.c.m. of two numbers =
If l.c.m.. of two numbers (greater than 1) is the product of them, then their g. c. d is ……
A. 1
B. 2
C. one of the numbers
D. a prime
g. c. d of two numbers =
=
If p1 and p2 are distinct primes, their g. c. d is
A. P1
B. P2
C. P1P2
D. 1
p1 and p2 are distinct prime numbers.
So, 1 is the only common factor among them.
Therefore, g. c. d of them is 1.
If p, q, r are distinct primes, their l.c.m.. is ……
A. pqr
B. pq
C. 1
D. pq + qr + pr
p,q and r are distinct primes.
So, they have no common factor other than 1.
Then their g. c. d is 1.
Therefore, l.c.m. of two numbers =
= = pqr
g. c. d (15, 24, 40) = …….
A. 40
B. 1
C. 1
D. 15 × 24 × 40
15 = 5 × 3 × 1
24 = 2 × 2 × 2 × 3 × 1
40 = 2 × 2 × 2 × 5 × 1
Therefore, g. c. d(15, 24, 40) = 1
l.c.m. (15, 24, 40) = …….
A. 1
B. 15 × 24 × 40
C. 120
D. 60
15 = 3 × 5
24 = 2 × 2 × 2 × 3
40 = 2 × 2 × 2 × 5
Therefore, l.c.m.(15, 24, 40) = 2 × 2 × 2 × 3 × 5 = 120
0. 02222... is a……
A. rational number
B. integer
C. irrational number
D. zero
It is repeating and terminating.
= ……
A.
B.
C.
D. does not exist
=
=
=
= …….
A. does not exist as a real number
B. does not exist as a binomial surd
C.
D.
=
Therefore, x = 9, y =
x2 – 4y = = 81 – 141 = – 60
x2 – 4y can not be the square of a rational number as we know that square of a rational number is always positive.
Therefore, the given expression does not exist as a binomial surd.
g. c. d (136, 221, 391) = …….
A. 136
B. 17
C. 221
D. 391
136 = 2 × 2 × 2 × 17
221 = 13 × 17
391 = 17 × 23
So, g. c. d. (136, 221, 391) = 17
l.c.m.. (136, 221, 391) = ………
A. 40664
B. 136 × 221 × 391
C. g. c. d(136, 221, 391)
D. 136 × 221
136 = 2 × 2 × 2 × 17
221 = 13 × 17
391 = 17 × 23
So, l.c.m.. (136, 221, 391) = 2 × 2 × 2 × 23 × 13 × 17 = 40664
If g. c. d (a, b) = 8, l.c.m.. (a, b) = 64 and a > b then a = ……..
A. 64
B. 8
C. 16
D. 32
We know that g. c. d. (a, b) × l.c.m.. (a, b) = ab
8 × 64 = ab
ab = 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Given, g. c. d (a, b) = 8 = 2 × 2 × 2
So, greatest divisor is 8.
Therefore, one of the number is 8.
Therefore the other number is
If g. c. d (a, b) = 1, then g. c. d (a — b, a b) = ………
A. 1 or 2
B. a or b
C. a + b or a— b
D. 4
Given g.c.d. (a,b) = 1
From Euclid’s algorithm we can say that if b|a, then g.c.d(a,b) = b
Let g. c. d (a—b, a + b) = k
Therefore we can say that k is a factor of both (a–b) and (a + b).
We can write a–b = rk for some r N
And also a + b = sk for some sN
Now, (a + b) + (a–b) = rk + sk
a + a + b–b = k(r + s)
2a = k(r + s)………..eq(1)
(a + b) – (a – b) = rk – sk
a + b–a + b = k(r–s)
2b = k(r–s)………..eq(2)
Also, g. c. d (a, b) = 1
Therefore, 2 × g. c. d (a, b) = 2 × 1
g. c. d (2a, 2b) = 2
g. c. d [(r + s)k, (r – s)k] = 2 (from eq (1) and eq (2))
k × g. c. d(r + s, r – s) = 2
= 2 × 1
So, k × g. c. d(r + s, r – s) = 2
k × g. c. d(r + s, r – s) = 2 × 1 = 2 × g.c.d(a,b)
By comparing we get k = 2
We know that 1 and 2 are co–prime numbers.
Similarly, we get k = 1
So, g. c. d (a—b, a + b) = k = 2
or g. c. d (a—b, a + b) = k = 1
g. c. d (a—b, a + b) = 1 or 2
1f n > 1, n4 + 4 is. n E N
A. a prime
B. a composite integer
C. 1
D. infinite
(as a2–b2 = (a + b)(a–b))
….. eq (1)
Now, n > 1
n – 1 > 0
Also,
Therefore, and are distinct positive integers.
So, we can say that
n–1 > 0
Similarly, n + 1 > 0
Therefore, are also distinct.
Thus, from eq(1) n4 + 4 has two distinct factors and 1 as a factor.
So, n4 + 4 is a composite number for n > 1.
If g. c. d (a, b) = 18, 1. c. m. (a, b) …….
A. 36
B. 72
C. 48
D. 108
Except 48 all the numbers are divisible by 18.
has …. digits after decimal point.
A. 5
B. 4
C. 3
D. 2
The decimal expansion of will terminate after …….. digits.
A. 4
B. 5
C. 3
D. 6
55 means the decimal expansion will terminate after 5 digits.
5n (n N) ends with ………..
A. 0
B. 5
C. 25
D. 10
Any power of 5 ends with 5.
2m5n (m, n N) ends with ………..
A. 0
B. 5
C. 25
D. 125
2m5n = 2m – 1 × 5n – 1 × 2 × 5 = 2m – 1 × 5n – 1 × 10
(m, n N so (m – 1), (n – 1) N – {0} )
2m × 5n ends with 0
represents
A. a terminating decimal
B. a non-recurring decimal
C. a recurring decimal
D. an integer
(5k + 1)2 leaves remainder ……….. on dividing by 5.
A. 2
B. 0
C. –1 or 1
D. 1
(5k + 1)2 = 25k2 + 10k + 1 = 5(5k2 + 2k) + 1
(5k + 1)2 leaves remainder 1 when divided by 5
On division by 6, a2 cannot leave remainder (a N)
A. 1
B. 4
C. 5
D. 3
From Euclid’s Division lemma, for any positive integer aN.
a = 6m + r where, 0<r<6 and m N
a = 6m, a = 6m + 1, a = 6m + 2, a = 6m + 3, a = 6m + 4 or a = 6m + 5
Now, a = 6m
Then a2 = 62m2
= 36m2 = 6 × (6m)2 + 0
Here, remainder is 0.
If a = 6m + 1
Then a2 = (6m + 1)2
= 36m2 + 12m + 1
= 6(6m2 + 2m) + 1
Here, remainder is 1.
If a = 6m + 2
Then a2 = (6m + 2)2
= 36m2 + 24m + 4
= 6(6m2 + 4m) + 4
Here, remainder is 4.
If a = 6m + 3
Then a2 = (6m + 3)2
= 36m2 + 32m + 9
= 6(6m2 + 6m + 1) + 3
Here, remainder is 3.
If a = 6m + 4
Then a2 = (6m + 4)2
= 36m2 + 48m + 16
= 6(6m2 + 8m + 2) + 4
Here, remainder is 4.
If a = 6m + 5
Then a2 = (6m + 5)2
= 36m2 + 60m + 25
= 6(6m2 + 10m + 4) + 1
Here, remainder is 1 = .
Thus in any case, if a2 is divided by 6, then the remainder is 0, 1, 3 or 4 but not 5.
Product of three consecutive integers is divisible by ………..
A. 24
B. 8 but not by 24
C. 6
D. 20
Product of three consecutive integers = odd even odd
Or Product of three consecutive integers = even odd even
In both the cases atleast one is even.
So, It is divisible by 2
Also, in three consecutive integers one of them is a multiple of 3.
Also, 2&3 are mutually prime numbers.
So, product of three consecutive integers is divisible by 6.