IIT JEE , AIEEE Forum - Mathematics , Algebra

umang

here are a few questions on binomial that are troubling me -

1. what is the remainder when -

(a) 4^96 is divided by 6 ?

(b) 2^2006 is divided by 17 ?

(c) 4(4!) + 5(5!) + ......20(20!) is divided by 64 ?

I really get confused while solving such type of questions !!

Pls suggest some shortcut ways to solve !!!

jatinroxx

Hey the last one's simple......

The nos. after 8(8!) divide 64.....

So u only have to consider the nos. b4 it.....

U can now do it by brute force......

nadeemoidu

1)
4^96 = 16 ^48 = (12 + 4) ^ 48
Expand by binomial theorem . all terms except 4^48 will be divisible by 6.

so the remainder will be same as that of 4^48 . continue this till u get a small no. ( here u 'll get 4^3 = 64 .

So the remainder will be 4.

2)
2^2006 = 4^ 1003 = 4 . 4^1002 = 4. 16^501 = 4. (17-1) ^501

Now apply binomial theorem .

feynmann

2 . apply Fermat's theorem

2 ^ 17 = 2 mod ( 17 )

raising both side to the power 118 , we get

2^ 2006 = 2^ 118 mod ( 17 )

multiplying both side by 2

2 ^ 2007 = 2 ^ 119 mod ( 17 )

= 2^7 mod ( 17 ) ( since 17 *7 = 119 )

or , 2 ^ 2006 = 2^ 6 mod ( 17 )

= 64 mod ( 17 )

= 13 mod ( 17 )

so reqd remainder is 13

hsbhatt

Last one 4.4!+5.5!+6.6!+...+20.20! = (5-1).4!+(6-1)*5!+...+(21-1).20!

=(5!-4!)+(6!-5!)+(7!-6!)+..+(21!-20!)

=21!-4!

Now [21/2] = 10 Hence 2¹⁰divides 21! which means 2⁸ also divides 21!

Hence 21!-4! = 64k-24 = 64k-24+64-64 =64k'+40. Hence the remainder on division by 64 = 40.

hsbhatt

Answer to Qn b)

Fermat's little theorem gives 2¹⁶-1 is divisible by 17 as 17 is a prime.

Hence 2¹⁶= 17k+1

Now 2006 = 16.125+6

Hence 2²⁰⁰⁶= (2¹⁶)¹²⁵*2⁶ = (17k+1)¹²⁵*64

= (17n+1)64 = 17m+64 = 17m+17*3+13

Hence the remainder is 13

The approach is similar to Feynmann's but the topic of congruences may not be familiar to all. The binomial theorem approach is easier to understand.

Knowing number theory up to Fermat's little theorem is unavoidable however.

akhil_o

For the first one...
remainder pattern of 4^n when divided by 6 is 4 4 4 4...
Hence the remainder has to be 4 for any power of 4

jatinroxx

Question (b)

Write the no. as

4* (17-1)^501

Expand this by binomial theorem.....

Leave all the terms with 17 in them...

Find the remainder of the remaining........

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