Hello, Ideasman!

This is can be solved with basic Algebra.

. . It took a while to construct a simple approach.

Also, there arefar too many wordsin the problem.

LetWhen a number of bananas is divided into 11 equal piles

. . with more than one banana in each pile, there are 6 bananas left over.

However, the bananas can be dividedexactlyinto 17 equal piles.

What is the smallest possible numbers of bananas?

Answer in back of book: 204Nbe the number of bananas.

Then we are told: . N .= .11a + 6

. . . . . . . . . and: . N .= .17b . . . . . for integersaandb.(a > 1)

So we have: .11a + 6 .= .17b

. . . . . . . . . . . . . .17b - 6 . . . . . .6(b - 1)

Solve fora: . a .= .--------- .= .b + ---------

. . . . . . . . . . . . . . . 11 . . . . . . . . . 11

Sinceais an integer, that fraction must also be an integer.

For a > 1, the least value occurs when b = 12.

Then: . a .= .12 + 6(11)/11 .= .18

Therefore: . N .= .11(18) + 6 .= .204