Question

a^m when divided by p gives a
remainder r and a^n when divided
by p gives a remainder s . show
that a^m+n when divided by p
will give remainder rs.​

Answer 1

Euclid's Division Lemma:

a = bq + r

Thus, for both the divisions, we get the equations as:

$\tt{a^{m} = pq + r}$

[The quotient is an unknown number. Thus, I let it as q for the first equation]

$\tt{a^{n} = pq' + s}$

[Let the quotient for the second equation be u]

In order to get $\tt{a^{m+n}}$, we need to multiply both the above equations. This is because we know the property of exponents:

When two exponents with same bases are multiplied, then the powers are added and the base remains same.

Thus:

$\tt{a^{m+n} = (pq+r)(pu+s)}$

=> $\tt{a^{m+n} = p^{2}qu + p(qs + ur) + rs}$

=> $\tt{a^{m+n} = p(pqu + qs + ur) + rs}$

Thus, if we compare this to the Euclid's Division Lemma we wrote in the starting, we get to know that:

a = $\tt{a^{m+n}}$

b = p

q = pqu + qs + ur

r = rs

Thus, the remainder is equal to rs.

Hence Proved.