 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Three Quotients and Integer Puzzle (Posted on 2015-06-03) For how many ordered pairs (p,q) of positive integers with q < p ≤ 2015 is each of p/q, (p+1)/(q+1) and (p+2)/(q+2) an integer?

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 4
We're given p=0 mod q, (p+1)=0 mod(q+1) and (p+2)=0 mod(q+2).

The last two congruences can be rewritten as p=q mod(q+1) and p=q mod(q+2) by adding the modulus value.

So we have (p-q) divisible by q*(q+1)*(q+2) and p will equal a multiple of that number plus q.  All that's left is to evaluate for appropriate q and count the multiples <= 2015.

q         q*(q+1)*(q+2)          multiples
1               6                            335
2        24                             83
3        60                             33
4           120                             16
5           210                               9
6           336                               5
7           504                               3
8           720                               2
9           990                               2
10        1320                              1
11        1716                              1

For example, given q=7, p=3*504+7=1519=7*219, 1520=8*190, 1521=9*169.

Thus there are 490 pairs (p,q) satisfying the problem conditions.

 Posted by xdog on 2015-06-03 11:41:28 Please log in:

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