Explanation: Unit digit in 7^{95} = Unit digit in (7^{4})^{23}\times 7^{3}
= Unit digit in 7^{3} (as unit digit in 7^{4}=1)
= Unit digit in 343
Unit digit in 3^{58} = Unit digit in (3^{4})^{4}\times 3^{2}
[as unit digit 3^{4}=1]
= Unit digit is 9
So, unit digit in (7^{95}-3^{58})
= Unit digit in (343 – 9) = Unit digit in 334 = 4
Unit digit in (7^{95}+3^{58}) = Unit digit in (343 + 9)
= Unit digit in 352 = 2
So, the product is 4\times 2=8

Explanation: Given, Total distance = 360 km
and, Speed of Sumeet = 12 km/h
Number of hours taken by Sumeet to complete 1 round = \frac{Distance}{Speed}=\frac{360}{12}= 30 h
and, Speed of John = 15 km/h
Number of hours taken by John to complete 1 round = \frac{Distance}{Speed}=\frac{360}{15}= 24 h
Thus, Sumeet and John complete 1 round in 30 h and 24 h, respectively.
Now, to find required hours, we find the LCM of 30 and 24.
⇒ 30=2\times 3\times 5
⇒ 24=2\times 2\times 2\times 3
Then, LCM (30, 24) = 2\times 2\times 2\times 3\times 5 = 120
Hence, Sumeet and John will meet each other again after 120 h.

Explanation: The prime factorisation of 240, 960 and 1024 is given below:
240=2\times 2\times 2\times 2\times 3\times 5 = 2^{4}\times 3\times 5,
960=2\times 2\times 2\times 2\times 2\times 2\times 3\times 5 = 2^{6}\times 3\times 5
and 1024=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 = 2^{10}
HCF of 240, 960 and 1024 = 2^{4}=16
Hence, there must be 16 books in each stack.
Now, number of stacks of Mathematics books =\frac{240}{16}=15
Number of stacks of Science books =\frac{960}{16}=60
and, number of stacks of Biology books =\frac{1024}{16}=64

Explanation: The number is short by 10 for complete division by 15, 25 or 35.

Explanation: We have, a+bp^{\frac{1}{3}}+cp^{\frac{2}{3}}=0 ...(1)
On multiplying both sides by p^{\frac{1}{3}}, we get
ap^{\frac{1}{3}}+bp^{\frac{1}{3}}\times p^{\frac{1}{3}}+cp^{\frac{2}{3}}\times p^{\frac{1}{3}}=0
⇒ ap^{\frac{1}{3}}+bp^{\frac{2}{3}}+cp^{\frac{3}{3}}=0
⇒ ap^{\frac{1}{3}}+bp^{\frac{2}{3}}+cp=0 ...(2)
On multiplying Eq. (1) by b and Eq. (2) by c, then
subtracting Eq. (2) from Eq. (1), we get
(ab+b^{2}p^{\frac{1}{3}}+bcp^{\frac{2}{3}})-(acp^{\frac{1}{3}}+bcp^{\frac{2}{3}}+c^{2}p)=0
⇒ (b^{2}-ac)p^{\frac{1}{3}}+ab-c^{2}p=0
[Here, p^{\frac{1}{3}} is an irrational number and ab-c^{2}p is a rational number]
Note that, sum of rational and irrational numbers cannot be zero.
So, (b^{2}-ac)p^{\frac{1}{3}}+ab-c^{2}p=0, only if
(b^{2}-ac)=0 and ab-c^{2}p=0
⇒ b^{2}=ac ...(3)
and ab=c^{2}p
⇒ a^{2}b^{2}=c^{4}p^{2} ...(4)
[squaring both sides]
From Eq. (4),
a^{2}(ac)=c^{4}p^{2} [since, b^{2}=ac]
a^{3}c-c^{4}p^{2}=0
⇒ c=0 or a^{3}-c^{3}p^{2}=0
⇒ c=0 or p^{2}c^{3}=a^{3}
On taking, p^{2}=\frac{a^{3}}{c^{3}}
We get, (p^{2})^{\frac{1}{3}}=\frac{a}{c} which is not possible as p^{\frac{2}{3}} is an irrational number and \frac{a}{c} is a rational number.
Hence, c=0
On putting c=0 in Eq. (3), we get
b=0
On putting b=0, c=0 in Eq. (1), we get
a=0
Hence, a=b=c

Explanation: 196=2^{2}\cdot 7^{2}, sum of exponents =2+2=4

Explanation: Sum is 888
⇒ unit’s digit should add up to 8. This is possible only for option (d) as “3”+“5” =“8”.

Explanation: Let, x=0.2\overline{54}, then
x = 0.2545454... ...(1)
Multiplying Eq. (1) by 100, we get
100x = 25.4545... ...(2)
Subtracting Eq. (1) from Eq. (2), we get
99x = 25.2
⇒ x=\frac{252}{990}=\frac{14}{55}
Comparing with \frac{p}{q}, we get
p=14
and q = 55
Hence, p+q = 14+55=69

Explanation: We know that, any odd positive integer is of the form 2q+1, where q is any integer. So, x=2m+1 and y=2n+1 for some integers m and n.
Now, x^{2}+y^{2}=(2m+1)^{2}+(2n+1)^{2}
= 4m^{2}+1^{2}+4m+4n^{2}+1^{2}+4n
= 4(m^{2}+n^{2})+4(m+n)+2
= 4[(m^{2}+n^{2})+(m+n)]+2
= 4r+2
Where, r=m^{2}+n^{2})+m+n
Clearly, 4r+2 is an even number and not divisible by 4.
Hence, x^{2}+y^{2} is even but not divisible by 4.

Explanation: Required number = H.C.F.{(245 – 5), (1029 – 5)}
= H.C.F. (240, 1024) = 16.

Explanation: L.C.M. \times H.C.F. = First number \times second number
Hence, required number = \frac{36\times 2}{18}=4

Explanation: In P, the primes that occurs are 2, 3, 5, 7.
In Q, there is no 2
So, the HCF of P Q , has powers of only 3, 5 and 7.
In P, 3 comes from 6, 12, 18. So, 3^{4} is the greatest power of 3 in prime factorisation of P . While in Q, 3 comes from 3, 9, 15.
So, 3^{4} is also in prime factorisation of Q.
Similarly 5^{2} is the greatest power of 5 for both P and Q and 7^{1} is the greatest power of 7 for both P and Q.
Hence, HCF of P and Q is (3^{3})(5^{2})(7).

Explanation: The L.C.M. of 16, 20 and 24 is 240. The least multiple of 240 that is a perfect square is 3600 and also we can easily eliminate choices (a) and (c) since they are not perfect number.

Explanation:
∵m=n^{2}-n=n(n – 1)
Now, m^{2}-2m=m(m – 2)
= n(n – 1)(n^{2}-n-2)
= n(n – 1)(n-2)(n+1)
Since we know that product of any four consecutive integers is always divisible by 24.
∴ m^{2}-2m is divisible by 24.

Explanation: p_1^2-p_2^2 is an even number.
Let us take p_{1}=5
and p_{2}=3
Then, p_1^2-p_2^2=25-9=16
16 is an even number.

Explanation: \frac{1}{7}=0.\overline{142857}
The second positive integer whose reciprocal have six different repeating decimals is
\frac{1}{13}=0.\overline{076923}
And the third positive integer whose reciprocal have six different repeating decimals is
\frac{1}{21}=0.\overline{047619}
Therefore, the values of x are 7, 13, 21
Hence, the required sum is = 7 + 13 + 21 = 41

Explanation: 0.1\overline{34}=\frac{134-1}{990}=\frac{133}{990}

Explanation: Let a_{1},a_{2},...,a_{100} be non-zero real number and
a_{1}+a_{2}+...+a_{100}
a_{i}2^{a_{i}}>a_{i} and a_{i}2^{-a_{i}}<a_{i}
∴ \sum_{i=1}^{100}a_{i}2^{a_{i}}>\sum_{i=1}^{100}a_{i} and \sum_{i=1}^{100}a_{i}2^{-a_{i}}<\sum_{i=1}^{100}a_{i}
⇒ \sum_{i=1}^{100}a_{i}2^{a_{i}}>0</span> and \sum_{i=1}^{100}a_{i}2^{-a_{i}}<0
Hence, option (a) is correct.

Explanation: Value of n = 2

Explanation: For n=1, 9^{n}=9^{1}=9, so 9^{n} can end with digit 9.
For n=2, 9^{n}=9^{2}=81, so 9^{n} can also end with digit 1.
Now, let if possible 9^{n} ends with 2 for some natural numbers n.
Then, 9^{n} is divisible by 2.
But prime factors of 9 are 3\times 3.
9^{n}=(3\times 3)^{n}=3^{2n}
Thus, prime factorisation of 9^{n} does not contain 2 as a factor.
By the fundamental theorem of arithmetic, there are no other primes in factorisation of 9^{n}.
Therefore, 9^{n} is not divisible by 2. So, our assumption is wrong.
Hence, there is no natural number n for which 9^{n} ends in the digit 2.

Explanation: 3125, 512 and 200 has factorization of the form 2^{m}\times 5^{n} (where m and n are whole numbers). So given fractions has terminating decimal expansion.

Explanation: H.C.F. of 20 and 15 = 5
So, 5 students are in each group.
∴ n=\frac{20}{15}=\frac{35}{5}=7
Hence, x = 4, y = 3 and n = 7

Explanation: Since n is divisible by 2 therefore (n+2) is divisible by 4, and hence n(n+2) is divisible by 8.
Also, n, (n+1), (n+2) are three consecutive numbers.
So, one of them is divisible by 3.
Hence, n(n+1)(n+2) must be divisible by 24.

Explanation: When 2^{256} is divided by 17 then,
\frac{2^{256}}{2^{4}+1}=\frac{(2^{2})^{64}}{2^{4}+1}
By remainder theorem when f(x) is divided by x+a the remainder =f(-a)
Here, f(a)=(2^{2})^{64} and x=2^{4} and a=1
Hence, Remainder =f(-1)=(-1)^{64}=1

Explanation: Let x=0.\overline{235} ...(i)
1000x=235.\overline{235} ...(ii)
Subtract (i) from (ii),
999x=235
⇒ x=\frac{235}{999}

Explanation: Since, H.C.F. of co-prime number is 1.
∴ Product of two co-prime numbers is equal to their L.C.M. So, L.C.M. = 117

Explanation: Given number is 1001. Then, the factor tree of 1001 is given as below

1001=7\times 11\times 13
By comparing with given factor tree, we get
x=7, y = 13

Explanation: Let x, x+36 and y are the three prime numbers.
According to the given condition,
x+x+36+y=100
⇒ 2x+y=64 ...(1)
Since x is a prime number, then 2x will be an even number.
Also addition of two even numbers results in an even number.
Hence, from equation (1), we can conclude that y must be an even prime number.
y = 2 as 2 is the only even prime number.
Put y = 2 in equation (1), we get
⇒ 2x+2 = 64
⇒ x = 31x = 31
So the required prime numbers are 31, 31+36, 2 or 31, 67 and 2
Hence, Largest number is 67.

Explanation: For all x∈N, (12)^{3^{x}} ends with either 8 or 2 and (18)^{3^{x}} ends with either 2 or 8.
If (12)^{3^{x}} ends with 8, then (18)^{3^{x}} ends with 2.
If (12)^{3^{x}} ends with 2, then (18)^{3^{x}} ends with 8.
Thus, (12)^{3^{x}}+(18)^{3^{x}} ends with 0 only.

Explanation: Given: The natural number, when divided by 13 leaves remainder 3
The natural number, when divided by 21 leaves remainder 11
So, 13 – 3 = 21 – 11 = 10 = k
Now, LCM (13, 21) = 273
But the number lies between 500 and 600
∴ 2 LCM (13, 21) – k = 546 – 10 = 536
536 = 19\times8+4
∴ remainder = 4

Explanation: For given numbers,
55^{725}, unit digit = 5;
73^{5810}, unit digit = 9
22^{853}, unit digit = 2
Unit digit in the expression 55^{725}+73^{5810}+22^{853} is 6

Explanation: H.C.F. (91, 126) = \frac{91\times 126}{L.C.M.(91,126)}=\frac{91\times 126}{182}=13

Explanation: Since, x is divided by 5, the remainder is 2
therefore x=5m+2
similarly, y=5n+4 consider x+y=5(m+n)+6
= 5(m + n) + 5 + 1
= 5(m + n + 1) + 1
But given that when x+y is divided by 5, the remainder is z
∴ z = 1
Now, \frac{2z-5}{3}=\frac{2(1)-5}{3}=-1

Explanation: Let us consider that n^{2}+3 is divisible by 17
∴ n^{2}+3=17K K ∈ N
⇒ n^{2}=17K-3
⇒ n^{2}=3(17m-1) [∵ K = 3m]
3(17m – 1) is a perfect square, which is not possible.
∴ n^{2}+3 is never divisible by 17.
In, n^{2}+4, put n = 9
So, (9)^{2}+4=81+4=85 which is divisible by 17.
∴ I is true and II is false.

Explanation: We have, X=28+(1\times 2\times 3 \times 4\times ...\times 16\times 28)
⇒ X=28[1+(1\times 2\times 3\times ... \times 16)]
Hence, X is a composite number.
Also, we have
Y=17+(1\times 2\times 3 \times 4\times ...\times 17)
⇒ Y=17[1+(1\times 2\times 3\times ... \times 16)]
Hence, Y is a composite number.
Now, X-Y=[1+(1\times 2\times 3\times ... \times 16)](28-17)
⇒ X-Y=[1+(1\times 2\times 3\times ... \times 16)](11)
Hence, X-Y is a composite number.
and, X+Y=[1+(1\times 2\times 3\times ... \times 16)](28+17)
⇒ X+Y=[1+(1\times 2\times 3\times ... \times 16)](45)
Hence, X+Y is a composite number.

Explanation: Let the numbers be 12x and 12y where x and y are co-primes.
Product of these numbers = 144xy
Hence, 144xy=6336
⇒ xy=44
Since, 44 can be written as the product of two factors in three ways. i.e. (1\times 44), (2\times 22), (4\times 11). As x and y are coprime, so (x,y), can be (1,44) or (4,11) but not (2,22).
Hence, two possible pairs exist.

Explanation: Required number = H.C.F. {(70 – 5), (125 – 8)}
= H.C.F. (65, 117) = 13.

Explanation: We have, \frac{51}{1500}=\frac{17}{500}
Prime factorization of 500=2\times 2\times 5\times 5\times 5 = 2^{2}\times 5^{3}
which is in the form 2^{n}\times 5^{m}
So, it has a terminating decimal expansion.
Now, \frac{51}{1500}=\frac{17}{2^{2}\times 5^{3}}
By comparing, we get n=2 and m=3
⇒ m+n=2+3=5

Explanation: 3^{13}-3^{10}=3^{10}(3^{3}-1)=3^{10}(26)=2\times 13\times 3^{10}
Hence, 3^{13}-3^{10} is divisible by 2, 3 and 13.

Explanation: All the given statements are false.

Explanation: We know that
13!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12\times 13
= 210\times 35\times 52\times 7\times 11\times 13
⇒ 24^{k}=(2^{3}\times 3)k
where k is largest non-negative integer
When 13! is an divided by 24^{k}, we get
\frac{2^{10}\times 3^{5}\times 5^{2}\times 7\times 11\times 13}{2^{3k}\cdot 3^{k}}
= 2^{10-3k}\times 3^{5-k}\times 5^{2}\times 7\times 11\times 13
∴ 10 – 3k is integer.
Then, maximum value of k = 3.

Explanation: For terminating decimal expansion, denominator must have only 2 or only 5 or 2 and 5 as factor.
Here, \frac{23}{8}=\frac{23}{2^{3}}
(only 2 as factor of denominator so terminating)