Show that 333^111+111^333 is divisible by 7

Asked by  | 25th Dec, 2008, 08:26: AM

Expert Answer:

Divisibility by 7.
Take both the terms one by one.

(111)^333= (7x15+6)^111

Now expanding the terms by using the binomial expansion.

(7x15+6)^111 = (7x15)^111 + A(7x15)^110 x (6)+ B(7x15)^109 x(6)^2 +……

Here all the terms are divisible by 7 (since there's a factor of 7 in all of them) except for the last one, which gives us 6^111. That's the only term that might not be divisible by 7. Let's find out what the remainder would be:
6=6

6x6= (7x5)+1=1

6x6x6= (7x30)+6=6

6x6x6x6= (7x186)+1=1

So it looks like every time we multiply by 6, the remainder flips between 1 and 6. If there are an odd number of 6s, then the remainder is 6. If there's an even number, then it's 1, so that tells us that 6^333=sx7+6.
so (111)^333=7s+6 for some number s.

Similarly:
(333)^111=(7x47+4)^111
Remainder=1^111=1
So 333^111=7t+1 and 111^333=7s+6, so
(333)^111+(111)^333=7t+1+7s+6
= 7(t+s+1), which tells us it is divisible by 7.


 

Answered by  | 2nd Jan, 2009, 03:17: PM

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