Question

dividing a number by 7 the remainder be 8/9/5​

Answer 1

Answer:

The smallest number which when divided by 7 or 8 leaves remainder 1 is 57.

Any number in the format of 56k+1 (for any k> 0 )will leave remainder 1.

let this number be N.

so, N=56k+1

if we divide N by 9 it should leave a remainder 5.

Therefore, we can say that (N−5)or(56k+1−5) is divisible by 9.

so, 56k−4 is divisible by 9

Now, this number should follow with the divisibility test of 9.

Adding up the digits we get, =5k+6k−4=11k−4

By hit and trial, for k=2 ,this turns out to be 11∗2−4=18 ,which is a multiple of 9.

By substituting k in N,

We get N=56∗2+1=113

Dividing 113 by 72 leaves remainder 41.