If in a long division sum,the dividend is 380606 and the successive remainders from the first to the last are 434, 125 and 413, then the divisor is?

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.

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

On dividing 1056 by 23, we get 21 as remainder
? Required number to be added = (23 - 21 ) = 2.

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.

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

Given Exp. = (a²+ b²+ ab) / (a³-b³)
= 1 / (a - b)
= 1 / (16 - 5)
= 1/11.

Let the number be N
According to the question 5N = 2N² - 3
? 2N² - 5N -3 = 0
? (N - 3) (2N + 1) = 0
N = 3 & -1/2
Thus the required number is 3

Let the number be N
According to the question,
[N x (3/2)] - [N / (3/2)] =10
? 3N/2 - 2N/3 = 10
? (9N - 4N)/6 = 10
? 5N / 6 = 10
? N = (10 x 6)/5 =12