Counting boxes and remainders

Question

You need to place the boxes in the cages so that in each cage there is the same amount of boxes. After placing 5, 6, 8 boxes, always one box was remaining. What is the minimum number of boxes there could be in this situation?

Answer 1

To have "always one box remaining," the minimum number of boxes must be the Least Common Multiple of 5, 6, and 8, ... PLUS 1.

As products of prime numbers,

    5 = 5

    6 = 2 * 3

    8 = 2 * 2 * 2

The LCM of 5, 6, and 8 is 2*2*2*3*5 = 120.   One more is 121.

Placing 5 boxes in each of 121 cages leaves 1 left over.

Placing 6 boxes in each of 121 cages leaves 1 left over.

Placing 8 boxes in each of 121 cages leaves 1 left over.

That is the situation described !!

Answer 2

Hi Gabrielius T.,

A fairly standard type SAT math question. The concept you need is "least common multiple". Look it up. Then you need the next integer higher than that.

In fact, I'll show you some of it.

Factor each of the numbers given (5, 6, and 8) into prime factors. The least common multiple is the product of those prime factors, using each the maximum number of times it shows up in any of those numbers.

So, you have: 1 x 5 , 2 x 3, 2 x 2 x 2 .

So you must multiply 2 x 2 x 2 x 3 x 5 = 120. Plus one = 121.

Why don't you try a few choices (much) less than that, to convince yourself that NO number smaller than 121 will meet the problem conditions. Then think for a while as to why this method works.

-- Cheers, --Mr. d.