Question

When 5655 is divided by a positive two digit intreger "N" the remainder is 11, when 5879 is divided by the same intreger N the remainder is 11.
What is the sum of the digits of N ?

Answer 1

Answer:

8

Step-by-step explanation:

Hi,

When we divide 5655 by N, we get remainder of 11, which means that 5655-11 is a multiple of N.

5655 - 11 = 5644 is a multiple of N.

Similarly, 5879-14 should be a multiple of N.

5879 - 14 = 5865 is a multiple of N.

Because 5644 and 5865 are both multiples of N, their difference must be a multiple of N.

5865 − 5644 = 221 then 221 is a multiple of N.

We have three number of which N can be a multiple of, however we choose to factorize the smallest possible number amongst these three, which is 221. (This is only for simplification of the solution, smaller the number, less the factors)

221 : 1, 13, 17, 221.

There are only two two - digit factors: 13 and 17.

We divide 5865 and 5644 by both numbers.

$\frac{5865}{13} = 451.15 \\\frac{5865}{17} = 345 \\\frac{5644}{13} = 434.15 \\\frac{5644}{17} = 332$

Looking at these results, we know only 17 divides all three numbers.

Hence N=17.

The sum of both digits will be: 1 + 7 = 8