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Divisibility by 7 (Posted on 2005-05-22) Difficulty: 2 of 5
(2222^5555 + 5555^2222) is or isn't divisible by 7 ?

Just pencil and paper.

  Submitted by pcbouhid    
Rating: 2.0000 (3 votes)
Solution: (Hide)

The answer is "IT IS", and we can prove it by several means. Here, a different way (a little insight and some knowledge of divisibility) from the correct answers given :

We have :

(2222^5555 + 5555^2222) = (2222^5555 + 4^5555) + (5555^2222 - 4^2222) - (4^5555 - 4^2222)

Let's analyse each sum enclosed by parentheses.

The first (2222^5555 + 4^5555) is divisible by (2222 + 4) = 2226 = 7 x 318, since (a^n + b^n) is divisible by (a + b) if n is odd, and so divisible by 7.

The second (5555^2222 - 4^2222) is divisible by (5555 - 4) = 5551 = 7 x 793, since (a^n - b^n) is always divisible by (a - b), and so divisible by 7.

The third (4^5555 - 4^2222) may be written :

4^2222(4^3333 - 1) = 4^2222(64^1111 - 1)

Since (64^1111 - 1) is divisible by (64 - 1) = 63 = 7 x 9 (see the second sum), so it is also divisible by 7.

So, (2222^5555 + 5555^2222) is divisible by 7.

Good work, everybody !!

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionAlternative MethodologyK Sengupta2007-11-14 04:12:54
re: Big numbers are NO PROBLEM for divisibilityphi2006-03-09 04:43:36
re(2): Big numbers are NO PROBLEM for divisibilityRupesh Khandelwal2005-05-25 15:43:38
Some Thoughtsre: Big numbers are NO PROBLEM for divisibilityRavi Raja2005-05-25 06:05:19
SolutionBig numbers are NO PROBLEM for divisibilityRupesh Khandelwal2005-05-25 05:52:18
Solutiondivisible by 7?heather2005-05-23 14:02:42
re: Divisible by 7Justin2005-05-22 17:02:50
SolutionDivisible by 7Ravi Raja2005-05-22 16:40:36
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