# Is it possible to solve a (simple) problem that includes remainders with basic algebra?

An eight-year old (Grade 3) told me about the "hardest problem" they had to solve on their math test yesterday. Here's the question:

There is a number less than 40, that when divided by 5 leaves a remainder of 3, and when divided by 6 leaves a remainder of 2.

He was very proud to have solved it himself :)

I immediately thought this might be a nice way to introduce him to simple algebra but when we reached home, I realised I couldn't come up with a way to express the problem.

My thoughts were along the lines of:

z = 5x + 3
z = 6y + 2


But then what can you do with

5x + 3 = 6y + 2


except for reaching

y = (5x + 1) / 6


and I have a feeling I'm going down a horribly wrong path.

Is there a "simple" way to solve this problem with algebra?

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It seems to me like your strategy is fine. In your last equation, you just evaluate at values of $x$ and see when you get an integer $y$. So $x=1, y = 1, z = 8$ will work. – Thomas Apr 5 '12 at 12:31
There are actually two numbers less than 40 (and greater than 0) that satisfy those equations. – Chris Taylor Apr 5 '12 at 12:35
Chinese Remainder Theorem! (Look up the...) – user1729 Apr 5 '12 at 12:43

When you write $y = (5x + 1) / 6$ your bad feeling is justified, because you could plug in an integer value for $x$ and get a non-integer value for $y$. Disaster!

One trick is to avoid any non-integer values by multiplying all coefficients of arbitrary values up to the least common multiple. For example, with

$$d = 5x + 3$$ $$d = 6y + 2$$

you have arbitrary numbers $x$ and $y$ in your solution, related through $d$. Their coefficients are $5$ and $6$, with least common multiple $30$, so if we multiply up:

$$6d = 30x + 18$$ $$5d = 30y + 10$$

you can subtract one from the other to get

$$d = 30(x-y) + 8$$

which is valid for all integer values of $x$ and $y$, so we may as well write them as a single variable:

$$d = 30m + 8$$

from which it is easily seen that 8, 38, 68, 98... are all solutions, and the ones between 0 and 40 are 8 and 38.

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This certainly seems much more straightforward to explain, although perhaps not quite at Grade 3.. at least I understand how to solve it now. I was hoping for something like your final line, but would have missed reducing the (x-y) to a single variable. Thanks! – Jedidja Apr 5 '12 at 13:27

$5x = 6y-1 = 6(y-1)+6-1 = 6(y-1)+5$. Hence $y-1$ is a multiple of $5$, $y=5t+1$, $x=6t+1$, and finally $z=30t+8$. If $z<40$, then $t=0$ and $t=1$ work and give $z=8$ and $z=38$.

The hard part is to conclude from $5x-5=6(y-1)$ that $y-1$ is a multiple of $5$. This is no longer algebra...

In the more general case, $z=5x+a=6y+b$, you can proceed as follows (which is different from the above ad-hoc solution): $6y+b=5x+a=6x-x+a$ and so 6 divides $x-a+b$ and so $x=6t+a-b$. The rest is similar.

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