Name: Tom
Level: Middle
Who: Uncle

My eighth grade niece called with the following homework problem:

A teacher wanted to divide her class into equally numbered groups. She tried to divide the class into groups of two, but was one student short. She tried to divide the class into groups of five, but was one student short. She tried to divide the class into groups of seven and was successful. What is the least number of students that were in her class?

I know the answer is 49, but don't know how to prove it. I must be getting old if I can't solve eighth grade math problems. Your assistance would be appreciated.

Tom,

There is a general method, called the Chinese Remainder Theorem, that solves such problems but at the Grade 8 level that's not want one wants. What you need to see is that

" She tried to divide the class into groups of two, but was one student short"

means that the number of students is somewhere in the sequence 1, 3, 5, 7, 9, ... (such a sequence, one where the difference between successive terms is always the same, is called an arithmetic progression)

"She tried to divide the class into groups of five, but was one student short"

means that the number of student is somewhere in the sequence 4, 9, 14, 19, ...

"She tried to divide the class into groups of seven and was successful"

means that the number of students is somewhere in the sequence 7, 14, 21, ... .

Thus you need to find the first place that the three sequences meet

	1,	3,	5,	7,	9,	...,	49,	...
	4,	9,	14,	19,	24,	...,	49,	...
	7,	14,	21,	28,	35,	...,	49,	...

I guess at this level the easy way to find the common point is to start down the sequence with the biggest differences (gaps), namely 7, 14, 21, ... and check if each number belongs to the other sequences. For example, it is easy to eliminate, 14, 28, etc., the even numbers in the last sequence, since the first sequence tells us we're looking for an odd number. Soon you will find 49 works.

Hope this helps!
Penny