What least number must be added to 1056 get a number exactly divisible by 23?

If sum of ODD positioned digits minus the sum of the EVEN positioned digits is divisible by 11 then the number is divisible by 11.

Sum of odd positioned digits - sum of even positioned digits = (9 + 2 + 5 + 6) - (7 + 1 + x) =14 - x

It must be divisible by 11. So x = 3.

6 + 1 + 3 + 5 + x + 2 = 17 + x must be divisible by 9.
So, x =1.

By hit and trial we find that 632 is divisible by 8. So, * must replaced by 3.

To determine if a number is divisible by 99 it needs to be divisible by 9 and 11.

Divisible by 9 test. If sum of digits is divisible by 9 then the number is divisible by 9.
Divisible by 11 test. If sum of ODD positioned digits minus the sum of the EVEN positioned digits is divisible by 11 then the number is divisible by 11.

Clearly 114345 is divisible by 9 as well as 11. so,it is divisible by 99.

Given Exp.= (1 - 1/3) (1 - 1/4) (1 - 1/5)...(1 - 1/n)
= (2/3) x (3/4) x (4/5) x ... x (n-1/n)
= 2/n.

For even values of n, the number (10ⁿ - 1 ) consists of even numbers of nines and hence it will be divisible by 11.

119 is divisible by 7, 187 is divisible by 11,247 is divisible by 13 and 551 is divisible by 19. So none of the given numbers is prime.

Let d = divisor and q = quotient

First remove the last remainder:
d x q + 413 = 380606
d x q = 380193

So d cannot be even (choice d) or a multiple of 5 (choice c).

If we try choice a or b,
a quick division (by calculator) shows that
380193 = 451 x 843, the other two choices.

(In the absence of choices, we could find the prime factors of 380193,
which are 3 x 11 x 41 x 281 and get various candidate divisors from that.)

So one of them is the divisor and the other the quotient.

So we can just check the first remainder:
843 x 4 = 3372
3806 - 3372 = 434 ... a match!

And just to be sure:
451 x 8 = 3608
3806 - 3608 = 198

So the divisor is 843.

Let the unit digits of the number be N and 10s digit be M
? Number = 10M + N

According to the question
MN = 10 .......(i)
And 10N + M = (10M + N) - 36
? 10N + M - 10M - N = -36
? 9N - 9M = - 36
? N - M = -4 .......(2)
On Adding eq. (i) and (ii) we get
N + M = 10
N - M = - 4
2N = 6
? N = 3 and M = 7
? Required product of two digits = 3 x 7 = 21

Let the five consecutive number are N - 2, N - 1, N, N +1, N+2
Sum of numbers = 190
? N-2 + N -1 + N + N + 2 = 190
? 5N = 190
? N = 38
Sum of largest and smallest numbers = (N + 2) + (N - 2)
= 2N = 2 x 38
= 76