Question

I want the answer to this question ASAP​

Answer 1

Note that 104 and 52 are multiples of 13. Then

100x+50+y=(104−4)x+(52−2)+y=104x+52+(y−2−4x).

So you need y−2−4x to be a multiple of 13. That is, you need y≡2+4x(mod13). So plug each possible value of x into the congruence and see if there is a solution for y in the range needed. There are at most, then 9 solutions.

Answer 2

Answer:

Hey friend!

The answer is option ( B)

There are 7 three digit numbers divisible by 13 having 5 as their middle number.

156, 351, 455, 559, 650, 754, and 858.

Hope this helps!

Have a good day ahead:)