Question

Anyone please answer question no.-1 and 2

Answer 1

1. By Euclid's division lemma,

A=bq+r where

$0 \leqslant r < b$

Since b = 3,

The only possible remainders are 0, 1 and 2 only

2. For a number to end with digit 5, it has to be a multiple of 5.

Prime factorisation of 6 = 2*3

${6}^{n} cannot \: end \: with \: digit \: 5 \: for \: any \: natural \: number \: n$

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