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# Test of Mathematics Solution Subjective 35 - Divisibility by 16

Test of Mathematics Solution Subjective 35 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## Problem

(a) Prove that, for any odd integer n, $n^4$ when divided by $16$ always leaves remainder $1$.

(b) Hence or otherwise show that we cannot find integers $n_1 , n_2 , ... , n_8$ such that $n_1^4 + n_2^4 + ... + n_8^4 = 1993$.

## Solution

For part (a) we consider $n = 2k +1$ and expand it's fourth power binomially to get $(2k+1)^4 = {{4} \choose {0}} (2k)^4 +{{4} \choose {1}} (2k)^3 + {{4} \choose {2}} (2k)^2 + {{4} \choose {3}} (2k)^1 + {{4} \choose {4}} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1$

Now $24k^2 + 8k = 8k(3k+1)$ ; if $k$ is even then $8k$ is divisible by $16$ and if $k$ is odd $3k+1$ is even and product of $8k$ and $3k+1$ is divisible by $16$. Since $16k^4 + 32 k^3$ is already divisible by $16$ we conclude $(2k+1)^4$ when divided by $16$ gives $1$ as remainder.

For part (b) we note that $1993$ when divided by $16$, produces $9$ as the remainder. Each of the eight of fourth powers when divided by $16$ produces either $0$ (when $n_i$ is even) or $1$ (when $n_i$ is odd using part (a)) as remainder. Thus they can add up to at most $8$ (modulo $16$) hence can never be equal to $9$ (which $1993$ is modulo $16$).

Test of Mathematics Solution Subjective 35 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

## Problem

(a) Prove that, for any odd integer n, $n^4$ when divided by $16$ always leaves remainder $1$.

(b) Hence or otherwise show that we cannot find integers $n_1 , n_2 , ... , n_8$ such that $n_1^4 + n_2^4 + ... + n_8^4 = 1993$.

## Solution

For part (a) we consider $n = 2k +1$ and expand it's fourth power binomially to get $(2k+1)^4 = {{4} \choose {0}} (2k)^4 +{{4} \choose {1}} (2k)^3 + {{4} \choose {2}} (2k)^2 + {{4} \choose {3}} (2k)^1 + {{4} \choose {4}} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1$

Now $24k^2 + 8k = 8k(3k+1)$ ; if $k$ is even then $8k$ is divisible by $16$ and if $k$ is odd $3k+1$ is even and product of $8k$ and $3k+1$ is divisible by $16$. Since $16k^4 + 32 k^3$ is already divisible by $16$ we conclude $(2k+1)^4$ when divided by $16$ gives $1$ as remainder.

For part (b) we note that $1993$ when divided by $16$, produces $9$ as the remainder. Each of the eight of fourth powers when divided by $16$ produces either $0$ (when $n_i$ is even) or $1$ (when $n_i$ is odd using part (a)) as remainder. Thus they can add up to at most $8$ (modulo $16$) hence can never be equal to $9$ (which $1993$ is modulo $16$).

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