Question

given "n" is the least possible integer such that the sum of the digits of "n" is 100 and the sum of the digits of "2∗n" is 110. Find "n".

Answer 1

N = 4999999999942

Sum of all digits of N is 100

2 * N = 9999999999884

Sum of all digits of (2 * N) is 110

Answer 2

Let the number be of 13 digits since we have to find a number whose sum of the digits  is 100.

So, we can take 9 nines( since maximum single digit is 9 and keeping the 9's back we can find the minimum number) and 3 other values whose sum of digits is 10 and sum of digits*2 = 20.

Then, we have, the number be 100*a + 10*b + c and we have to find the minimum number such that

(a, b, c)= (2, 4, 4) [Since, 9*2 = 18 and we have to keep "8" in the digit]

So, we have a multiple number of n's.

n= 2449999999999 and 2*n = 4899999999998...........................(1).

                                     or,

n= 2499999999994 and 2*n = 4999999999988............................(2)

                                    or,

n= 4999999999924 and 2*n = 9999999999848............................(3)

In this way, we can get 6 combination of numbers whose sum of the digits is 100 and sum of the digits multiply by 2 is 110.

minimum of them = n= 2449999999999

ANSWER: 2449999999999