How Cheenta works to ensure student success?
Explore the Back-Story

Test of Mathematics Solution Subjective 35 - Divisibility by 16

Test of Mathematics at the 10+2 Level

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 at the 10+2 Level

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

Your email address will not be published. Required fields are marked *

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

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.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
Math Olympiad Program
magic-wandrockethighlight