# Blog

## The 63 dollar question

By

,

On Dr Vine’s desk there is some money.

• There are six pieces of Australian money.
• The money adds up to \$63.
• There are no \$1 coins.
• Each note is different from the others.

Can you work out what denominations are on the desk?

First, the answer: a \$50 note, a \$5 note and four \$2 coins will do it!

So how do we find the answer? First, you’re going to need a \$50. Without it, there’s no way you can get close to \$63 without having multiple notes with the same value.

Now, you need \$13 from the remaining five pieces of money. There’s no way to do that with coins – five \$2 coins is only \$10. So you’ll need one more note, which could be either \$5 or \$10.

Let’s try a \$10 first. With a \$50 and a \$10 on the desk, there are four more pieces of money, they add to \$3, and none of them can be \$1 coins. Four 50c coins only add to \$2, and if you try using a \$2 coin instead, you still won’t be able to find an answer.

If you try the \$5 note, things work much better. Then the four remaining pieces of money add to \$8, and that’s easy: four \$2 coins!

For more brainteasers and puzzles for kids, subscribe to Double Helix magazine!

Categories:

## 4 responses

1. Martin

I worked this out in an almost reverse order.

I noticed that \$63 is an odd number and not allowing \$1 means we need to get it to an even number somehow and the only way to do that is with a \$5 note, leaving \$58 behind.

I didn’t like the \$50 part so I took care of that with a \$50 note, leaving \$8.

The easiest ways to divide that up is to either use [\$1, \$2, \$5] (which isn’t allowed because of the \$1) or [4 x \$2]

Giving: \$50, \$5, \$2, \$2, \$2, \$2

2. David

Nice!

You could also get an odd number of dollars with two 50 cent coins, right?

Still, it’s probably a fair assumption that the answer probably doesn’t use anything smaller than a dollar.

It’s quite fun trying to rewrite the puzzle for different currencies. It was much easier to phrase back in 1973, for instance – you can replace the ‘every note is different’ rule with ‘all six pieces of money are notes’!

3. Mark W

What about: \$50 + \$5 + \$5 + \$2 + .50 + .50 ?

1. David

It adds up to \$63, but it doesn’t work with the last rule:
Each note is different from the others.

Although thinking about it, you could always use a \$5 coin…

This site uses Akismet to reduce spam. Learn how your comment data is processed.