Numbers and Divisibility

Sum of odd place digits $=2+1+2=5$
Sum of even place-digits $=8+5+3=16$
Difference of above two sums $=16-5=11$
The difference 11 is divisible by 11
$\therefore$ 3,25,182 is also divisible by 11.

These numbers are divisible by 5 and 7..

Multiplying unit digit of each number we get 672 hence 2 is the unit digit of the product.

The product (n-1)! is divisible by n if n is composite and n > 4 (Wilson's Theorem related logic). For n in [12, 40], (n-1)! is not divisible by n only if n is prime. The primes between 12 and 40 are 13, 17, 19, 23, 29, 31, 37. There are 7 such numbers.

On dividing we get,
$\frac { 9217 }{ 88 } =104\frac { 65 }{ 88 } $

Therefore,

Required number 

$= 9217 + (88 - 65)$ ,Because (88 - 65) < 65.

$= 9217 + 23$

$= 9240$

803642 will be divided by 9, if 8+0+3+6+4+2 =23 is divided by 9.
$\Rightarrow$ needs 4 more for that.

The largest 4-digit number is 9999. Dividing 9999 by 88 gives 113 with a remainder of 55. Subtracting 55 from 9999 yields 9944.

This is a duplicate of 520523. The largest 4-digit number divisible by 88 is 9944.

The least number divisible by 12, 15, 20, and 27 is their Least Common Multiple (LCM). The prime factorizations are 12=2^2*3, 15=3*5, 20=2^2*5, and 27=3^3; the LCM is 2^2*3^3*5 = 4*27*5 = 540.

To find the unit digit of the product, multiply the unit digits: 4 * 8 * 7 * 3. 4 * 8 = 32 (unit digit 2), 2 * 7 = 14 (unit digit 4), 4 * 3 = 12 (unit digit 2).

A number is divisible by 9 if the sum of its digits is divisible by 9. The sum of digits of 5x311 is 5+x+3+1+1 = 10+x. For 10+x to be divisible by 9, x must be 8 (18 is divisible by 9).

A number is divisible by 3 if the sum of its digits is divisible by 3. Sum = 1+8+a+7+1 = 17+a. For 17+a to be divisible by 3, a must be 1 (18) or 4 (21) or 7 (24). Given options, 1 is correct.