Show that 333^111+111^333 is divisible by 7
Asked by | 25th Dec, 2008, 08:26: AM
Divisibility by 7.
Take both the terms one by one.
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:
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.
So 333^111=7t+1 and 111^333=7s+6, so
= 7(t+s+1), which tells us it is divisible by 7.
Answered by | 2nd Jan, 2009, 03:17: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number