Given a pair of positive integers

*a*,

*b*, the operation

*a*∗

*b*is defined as 23 ×

*a*+ 31 ×

*b*.

For example, 20∗15 = 23 × 20 + 31 × 15 = 925.

Determine both pairs of positive integers

*x*,

*y*such that

*x*∗

*y*= 2015.

For me this problem is just a regular Diophantine equation. So I find the standard form (via Euclid's method or just checking some of the simple combinations)

3 * 31 - 4 * 23 = 1

From there you multiple everything by 2015:

6045 * 31 - 8060* 23 = 2015

which gets you a solution but not with 2 positive integers.

You can then adjust both terms by adding and subtracting 31 * 23 or 713 to find other solutions

So add and subtract 261 * 23 * 31 == 8091 * 23 == 6003 * 31

and you get

42 * 31 + 31 * 23 = 2015 and you're found the first pair.

Scale down once again to

19 * 31 + 62 * 23 and you have the next pair

As you can see the next scaling will cause the first term to go negative so there really are only 2 pairs of positive tuples.

However none of the kids know number theory

**so the question is whether this is really solvable for them?**

One thought is that you could notice that 2015 is divisible by 31

31 * 65 = 2015 and then perhaps stumble onto scaling to go down to the 2 solutions. From there if we explored scaling a bit with some easier equations first this might be more obvious. Alternatively, I could just ask them to find the standard form via trial and error and go from there.

I'm, still brainstorming if I can think a sequence of problems to bridge this gap.

To provide some background, this question was posed to Y6 students on an exam. The sole topic tag on our blog that links to this problem is "factorisation", which is a hint to a workable approach for that age group.

ReplyDeleteThat would be an entry to finding its divisible by 31.

ReplyDeleteI think you could approach with the "table" route, making guesses and refining them. Some numbers too big, some too small. Because a and b are integers, you're not dealing with an uncountable number of choices.

ReplyDelete