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## How many hats? Brainteaser

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Difficulty: Extreme!

The geology club was on its way to check out a fault line. As the minibus cruised down the highway, several club members realised they had forgotten their hats.

The bus pulled in at the next petrol station, and the teacher bought a bunch of hats. The day was saved!

A few weeks later, the teacher was looking through their credit card bill, trying to remember what they bought. They spent \$131 on hats.

One of the students took some funny photos in the petrol station, and the teacher could just make out the prices:

• \$20 for a cap
• \$21 for a bucket hat
• \$24 for a straw hat.

Can you work out how many of each hat the teacher bought?

Scroll down or click for a hint, or the answer!

## Brainteaser hint

A good first step is to work out how many hats they bought in total.

First, it’s a good idea to work out the total number of hats. One way to work it out is by guessing and checking. Six hats seems like a good guess. To check:

• If the teacher bought 7 of the cheapest hat, it would still be more than \$131.
• If they bought 5 of the most expensive hat, it would still be less than \$131.
• So, there must be 6 hats.

All the hats have similar prices. We can use this to our advantage. Imagine the cap as \$20 + \$0, the bucket hat as \$20 + \$1, and the straw hat as \$20 + \$4.

We can write the total as:
\$20 + \$20 + \$20 + \$20 + \$20 + \$20 + some \$0s, \$1s and \$4s = \$120 + \$11

If we remove \$120 from each side, we get:
Some \$0s, \$1s and \$4s = \$11

Since we only have 6 hats, 11 x \$1 isn’t a solution. Neither is 1 x \$4 + 7 x \$1. The only solution that works is:
\$0 + \$1 + \$1 + \$1 + \$4 + \$4 = \$11

And if we add the \$20s back in to check, we get:
\$20 + \$21 + \$21 + \$21 + \$24 + \$24 = \$131

The teacher bought one cap, three bucket hats and two straw hats.

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## 3 responses

1. Peter

The math in my head was pretty much the same, though seemed simpler.
131=20×6 with 11 remainder
11=4×2 with 3 remainder
6-(3+2)=1
So 3 bucket hats at \$21 each, 2 straw hats at \$24 each, and 1 cap at \$20.

2. Nathan Duhig

I was less systematic/algebraic, more guess, check and refine.
I looked at the one dollar of \$131 and first tried a bucket hat, subtracting \$21 leaving \$110. There was no way I could combine any other hats to make \$110.
My next thought was to make up \$11 from 2X\$4 (straw hats) and 3x\$1 (bucket hat). The total cost of these 5 hats came to \$111, leaving room for just one more cap to make up the total.

3. Mike

My train of thought was similar to Nathan’s. The total is odd and the only odd priced hat is the bucket hat. So there should be an odd number of bucket hats.
7X21=147 too much.
5X21=105 and 131-105=26 no hat has this price
1X21=21 and 131-21=110. There is no combination of 20 and 24 that add up to 110, so the only option left is
3X21=63 and 131-63=68. The last digit (8) indicates two straw hats. 2X24=48 and 68-48=20 so there is only one cap.

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