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let nn denote the smallest positive integer that is divisible by both 44 and 9,9, and whose base-1010 representation consists of only 44's and 99's, with at least one of each. what are the last four digits of nn?

Respuesta :

The last four digits of the number nn are 4944

Given that,

Let nn stand for the lowest positive integer that can be represented in base-1010 using only the numbers 44 and 99, with at least one of each. This number must be divisible by both 44 and 9,9.

The last two digits of n must be divisible by 4 for n to be divisible by 4.

So last digits of n will be 44.

And for divisiblity by 9, sum of all digits of n has to be divisible by 9.

As two digits are fixed and their total is 8, next feasible total will be 45 so 9 can be atleast one time in n.

And as statement says n has to be smallest number n will be 4444444944

So last 4 digits of n are 4944.

Therefore, nn's last four numbers are 4944.

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