Date: Wed, 06 Nov 1996 18:26:15 -0500

From: Alex Bogomolny

Dear Richard:

judging by the answer you provide your problem should read

> Find four consecutive integers such that the first and third is five

> more than the fourth.

This is the problem I am going to solve. Denote the smallest number by a. Then we get four numbers

a, a+1, a+2, a+3.

The sum of the first and the third equals a+(a+2) which is 2a+2. This is 5 more than the last number, i.e.

(2a+2) - 5 = a+3

This gives 2a+2-5=a+3 or 2a-3=a+3 and, finally, 2a-a=3+3. a=6. That's it. The power of algebra.

Regards

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