# A Countdown Conundrum

Not *that* sort of conundrum, obviously.

Like everyone else on the planet, and presumably others, I got sucked into the craze of Wordle and its spinoffs during the pandemic. The one I turn to first? Countle.

It’s the number games from Countdown, about which I’ve written before here and at Chalkdust. You’re given six numbers chosen from a list – 1 to 10 appear twice, and the four big numbers (25, 50, 75 and 100) once each – and a target you have to reach.

You may use each number at most once; you may not use non-positive integers; you may use only the four basic arithmetic operations.

Today (I’m writing this in October 2022), the numbers were 25, 50, 75, 100, 4 and 6; the target was 986. Have a go, if you’d like. I’ll spoiler the solution after the line.

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I tried many blind alleys before stumbling on the following solution:

- $75 + 4 = 79$
- $79 \times 50 = 3,950$
- $3,950-6 = 3,944$
- $100 \div 25 = 4$
- $3,944 \div 4 = 986$.

Goodness. Obviously, I didn’t manage that within the allowed 30 seconds.

What I wanted to explore was, what sort of thought process gets you to that solution?

986 is an awkward number, sort of midway between 975 and 1000, too far from either to reach easily. That “midway” piece is key.

Rather than looking at 39 or 40 twenty-fives, we need to look at 79 twelve-and-a-halves, which is 987.5, which is one and a half away. Obviously, we’re not allowed halves, so we’d need to go up higher and back down again. Our options would be something like:

- $79 \times 25 \div 2$
- $79 \times 50 \div 4$
- $79 \times 75 \div 6$
- $79 \times 100 \div 8$.

The middle two look more likely, since we have a four and a six available. But there’s a problem with each: the first requires two fours (because 79 is most likely to come from 75 + 4) and the second requires two 75s (for much the same reason).

At this point, the 1.5 discrepancy comes into play: if we’re going to be dividing by 4, we could remove 1.5 by subtracting 6 before dividing. In the six case, we could subtract 9 – the first looks more likely, because we do have a six.

And we can *get* a second 4 from $100 \div 25$, neither of which we’ve used.

I’m not 100% satisfied with this explanation – and it’s absolutely not how I tackled it – but it feels like a useful tactic.

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My actual approach, which I characterised as trial and error, did have some method to it in the end.

- I had worked out that 986 is a bit short of a quarter of 4,000,
- I knew that $75 \times 50$ is 3,750.
- A quarter of that is 937.5, so I’d need an extra 200 or so.
- I could get that from adding four to the 75 before multiplying – so $79 \times 50$ is 3,950.
- That seems pretty close, so I figured out $986 \times 4 = 3,944$
- And that’s just six away from where I am – I can get there!

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Do you have a different way – either of solving it or thinking about it? I’d love to learn your ways!