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Doubtful Divisibility (Posted on 2012-05-09) Difficulty: 1 of 5
For two positive integers x and y, we have:
(100*x + y)/(x+y) = 1 + (99*x)/(x+y) and,
(100*x + y)/(x+y) = 100-(99*y)/(x+y) . . . . (i)

Since both x and y are positive, we have: x+y > x and x+y > y

Accordingly, x+y does not divide either x or y, and, consequently from (i) we deduce that x+y divides (100*x + y) whenever, (x+y) divides 99.

Now, substituting (x, y) = (20, 16), we have: x+y = 36 which does not divide 99 and consequently we can conclude that 36 does not divide 2016.

However, in reality 2016/36 = 56 and, accordingly the above result is flawed.

Can you spot the error?

No Solution Yet Submitted by K Sengupta    
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Comments: ( Back to comment list | You must be logged in to post comments.)
Error(s) spotted | Comment 1 of 3
"consequently from (i) we deduce that x+y divides (100*x + y) whenever, (x+y) divides 99."

This is not true.  It should say x+y divides (100*x + y) whenever, (x+y) divides 99*x.

Which in your example it does:  99*20 = 1980 and 1980/36 = 55

(I also just noticed a punctuation error:  there should not be a comma after the word whenever.)

  Posted by Jer on 2012-05-09 12:17:47
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