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.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

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.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

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 $16$ ).

Leave a Reply Cancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

HALL OF FAME BLOG OFFLINE CLASS

CHEENTA ON DEMAND BOSE OLYMPIAD

Cheenta Academy Private Limited |

support@cheenta.com

Menu

Math Olympiad Program