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 1248 must be included (Posted on 2016-09-06)
There are three infinite arithmetic progressions.
Given that each of the numbers 1,2,3,4,5,6,7,8 occurs at least in one of these progressions, show that the number 1248 must occur at least in one of them.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Analytical solution | Comment 2 of 3 |
Assume that no series has 1248 in it.

1248 - 8 = 1240, which is a multiple of 1,2,4 and 5.
So, whichever series has 8 in it cannot have 3 or 4 or 6 or 7, because those would give rise to differences of 1 or 2 or 4 or 5.

1248 - 6 = 1242, which is a multiple of 1,2,3 and 6.
So, whichever series has 6 in it (and it cannot be the series with 8 in it) cannot have 3 or 4 or 5 or 7 or 8.

So 4 must be in the third series (the one without 6 and without 8).
But 3 cannot be in that series either, because otherwise all integers greater than 3 are in the series.

Therefore, 3 is not in any of the three series, which is a contradiction.

So the initial assumption is false, and 1248 is in one of the series.

 Posted by Steve Herman on 2016-09-06 10:54:23

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