Answer: Option C. -> 0
:
C
option (c)
Any power of 10 when divided by 3 gives a remainder of 1. Also,7 divided by 3 remainder is 1. Therefore individual remainders are 1 and 1.
Thus, 1-1=0. Answer is 0

Answer: Option B. -> II,III and IV
:
B
option(b)
take $N=3$ as $\frac{3}{15}$ gives a remainder 3.
$6N=18\frac{18}{15}$ does not give a remainder 6
$7N=\frac{21}{15}$ remainder =6
$2N=\frac{6}{15}.$ remainder =6
$5N+6=\frac{21}{15}.$ Remainder = 6

Answer: Option C. -> 375
:
C
option (c)
Numbers which cannot be expressed as difference of squares are all even non-multiples of 4. There are 250 even non-multiples of 4 from 1 to 1000 and 125 such numbers between 1 and 500 (2,6,10......). 125 cannot be expressed hence 375 can be expressed. Hence option c

Answer: Option E. -> 1
:
E
If a number is divisible by 72, it is divisible by 8 and 9. To be divisible by 8, the number formed by the last three digits should be divisible by 8. Therefore, 31Y should be divisible by 8$\Rightarrow$ Y = 2. To be divisible by 9, the sum of the digits of the number should be divisible by 9$\Rightarrow$ X = 3. Therefore, X-Y = 1

Answer: Option A. -> 240
:
A
Note: This is based on the concept of Euler’s number
A number N can be written as $a^m{b}^n$. In this case, $N=900=2^2\times 3^2\times 5^2$
The question can be rephrased as
The number of natural numbers which are less than 900 and relatively prime to it is
$N=[1-\left(\frac{1}{a}\right)][1-\left(\frac{1}{b}\right)][1-\left(\frac{1}{c}\right)]$
$N=[1-\left(\frac{1}{2}\right)][1-\left(\frac{1}{3}\right)][1-\left(\frac{1}{5}\right)]=240$

Answer: Option C. -> -1
:
C
32540 using the divisibility test for 13 ( triplets at odd- triplets at even places ) can be written as
540-32= 508. 508 when divided by 13 can be represented as 13k + 1 (i.e. when the number is divided by 13, the remainder is 1)
Similarly
42400= 400-42= 358 = 13k + 7
3233= 233-3=230= 13k + 9
(13k + 1)( 13k + 7)(13k + 9) will give the non-divisible part as $(9\times 7\times 1)=63$
therefore 63+222= remainder of 285 when divided by 13= -1 or 12

Answer: Option C. -> 340
:
C
Write all the numbers in base $10$ and then arrive at the answer
$127$ in base $8=7\times 1+2\times 8+1\times 8^2=87$ in base $10$
$12$ in base $6=2\times 1+1\times 6=8$ in base $10$
Adding both $=95$ in base $10$
Expressing in base $5=340$

Answer: Option B. -> 1363
:
B
Following are the steps while calculating the answer
4 digit numbers divisible by $11=\frac{8999}{11}=819,4$ digit numbers divisible by $13=\frac{8999}{13}=692$ or $693$
Number of integers common to 11 and 13 (as questionis 11 or 13) $=\frac{8999}{(13\times 11)}=62$ or $63$
To subtract numbers which are common to $11\&17;13\&17;11,13\&17$
$11\&17=\frac{8999}{(11\times 17)}=47$ or $48$ terms
$13\&17=\frac{8999}{(13\times 17)}=40$ or $41$
$11,13,17=\frac{8999}{(11\times 13\times 17)}=3$ or $4$ terms
Total ( you can approximate as the answers are far apart)= 819+693-63-48-41+4= 1364
Closest answer is option (2)

Answer: Option D. -> 0
:
D
Note that 71! has 16 zeroes at its end.(You can find this out by finding the highest power of 5 in 71!)
So does 70!.
66! To 69! Have 15 zeroes at its end. Multiplication of these itself amounts to $>202$
Hence the ${202}^{nd}$ digit is 0
To find the highest power of a number in a factorial
a)Highest power of a prime number in a factorial:
To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.
Eg) Find the highest power of 100!
Solution:
$\frac{100}{5}=20;\frac{20}{5}=4;$
Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24
b)Highest number of a composite number in factorial
1)Factorize the number into primes.
2)Find the highest power of all the prime numbers in that factorial using the previous method.
3)Take the least power.

Answer: Option C. -> 04
:
C
After 10!, the factorials end with 0 as the last two digits. We thusneed to find the last two digits from 4! To 9! Only
4!+5!+6!....9!= 24+120 + 720 + 5040 + 40320 + 362880 = ___04