2021-09-07 Year 1 viral maths problem

From Wikistix


Karla says, "I have 3 hundreds counters, 17 tens counters and 16 ones counters."

  1. Can she make two equal three-digit numbers? If so, draw the counters to show them.
  2. Can she make two equal three-digit numbers if she has to use all her counters? If so, draw the counters to show them.


Part a. is easy, we're not forced to use all the counters, so many combinations are possible. Perhaps the simplest is to split each set of counters in half, resulting in:

  • Each gets 1 hundreds counter, 8 tens counters and 8 ones counters, totalling 188.
  • 1 hundreds counter and 1 tens counter remaining.

The more interesting question is part b. Can we use all the counters and still have equal totals? The first observation that simplifies this is that if we were to have all 3 hundreds counters on one side, the remaining counters only add up to 186, so the hundreds counters must be distributed, 1 and 2.

This turns the problem into an equality with 2 unknowns, and we can use some simple algebra:

[math]\displaystyle{ a = \text{number of tens counters on one side} }[/math]
[math]\displaystyle{ b = \text{number of ones counters on one side} }[/math]

Now form the equality, using [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] on one side, and expressions for the remainder on the other side.

[math]\displaystyle{ \begin{align} 200 + 10a + b & = 100 + 10(17 - a) + (16 - b)\\ 200 + 10a + b & = 100 + 170 - 10a + 16 - b\\ 200 + 10a + b & = 286 - 10a - b\\ 20a + 2b & = 86\\ 10a + b & = 43 \end{align} }[/math]

At this point, we can see that either:
[math]\displaystyle{ a = 4, b = 3 }[/math]
[math]\displaystyle{ a = 3, b = 13 }[/math]
In either case, each side totals 243.

So, that's using Year 7 or Year 8 algebra. How is a seven year old (Years 1 or 2) supposed to solve this? I would suspect that trial and error is expected, and the way to go. After figuring out how to split the hundreds, splitting the tens and then ones should not take that much time - indeed, it's likely faster than doing the mechanics above!

Misinformation found herein copyright Paul Ripke (aka “stix”) stixpjr@gmail.com.