   SEARCH HOME Math Central Quandaries & Queries  Question from Ingrid, a parent: I am trying to help my son with an area question. I have the answer, from the solutions, but I cannot figure out how to teach him. Question: Two rectangles have lengths 13cm and 19cm respectively. Their total area is 376cm squared. If both their widths are whole numbers, what is the difference in their areas? I know that this is solvable once I determine the widths of the rectangles , but how do I go about finding that? Thanks for your help Hi Ingrid,

At the grade 6 level the best approach is probably the following.

You want

an integer multiple of 19 + an integer multiple of 13 = 376

which you can restate as

an integer multiple of 13 = 376 - an integer multiple of 19.

The procedure is then to take integer multiples of 19, subtract from 376 and see if the difference is divisible by 13. To do this in an orderly way you might use a table. Lets call $w$ the integer you multiply by 19.

w 376 - 19w Divisible by 13?
1 $376 - 19 \times 1= 357$ No!
2 $376 - 19 \times 2 =338$ Yes. $338 = 13 \times 26$

Thus two rectangles that satisfy the conditions in the problem are one that is 19 cm by 2 cm and the other being 13 cm by 26 cm.

To see if this is the only answer you could continue the table.

w 376 - 19w Divisible by 13?
1 $376 - 19 \times 1= 357$ No!
2 $376 - 19 \times 2 =338$ Yes. $338 = 13 \times 26$
3 $376 - 19 \times 3 =319$ No!
4 .... ....

If you continue to the row with $w = 15$ you will see that there is a second answer.

If this problem intrigues you we suggest that you read about Sylvester’s problem.

(If a and b are relatively prime natural numbers then every integer greater than (a-1)*(b-1) can be written as a nonnegative linear combination of a and b.)

We hope this helps,
Chris and Harley      Math Central is supported by the University of Regina and the Imperial Oil Foundation.