Solution to Luke's problem

The problem I was set by my grandson Luke was this:

"Find a 5-digit number which when multiplied by 4 gives a 5-digit result with the digits of the original number in the reverse order."

You should see straight off that the first digit in the number can be only 1 or 2. If it is 3 or more then multiplying it by 4 would give a 6-digit answer (for example, 30000 × 4 = 120000).

You can then rule out 1 as the first digit, because it would also have to be the final digit of the answer: but the answer after multiplying by 4 must be even! So we have got the first digit of our 5-digit number – and the final digit of the result of multiplying it by 4 – in both cases it is 2.

Now this means that the final digit of the starting number must produce a result ending in 2 when multiplied by 4. So it is either 3 (3 × 4 = 12) or 8 (8 × 4 = 32).

So is our number 2abc3 or 2abc8? (using a, b and c for for the missing digits)

Now it's pretty obvious that 2abc3 × 4 cannot equal 3cba2. So our number must be 2abc8.

Now 2abc8 × 4 = 8cba2 looks a real possibility, because the leading digit 2 on the left when multiplied by 4 gives the 8 needed in the result. But we will get 8 only if there is no 'carrying' from the result of multiplying a by 4. So, a can be only 0, 1 or 2.

So we have three possibilities now:

20bc8 × 4 = 8cb02

21bc8 × 4 = 8cb12

22bc8 × 4 = 8cb22

It's got to be the middle one! Why?

(Tricky to explain, but I'll have a go. In the first case, the 0 would have to be the final digit in the result when you multiply c by 4 and add the 3 carried from 8 × 4 = 32. This is impossible because c × 4 must be even! The same argument applies if the second digit in the number is 2.)

So, I now know the number is 21bc8, and that 21bc8 × 4 has to equal 8cb12.

To get that 1 in the answer 8cb12, then c × 4 must end in an 8; hence c is either 2 or 7.

So, I now have two possibilites:

21b18 × 4 = 81b12

21b78 × 4 = 87b12.

If you start multiplying 21b18 by 4, working from the units digit first, you find the answer must end in 72. So, that's no good.

So, we now know our number must be 21b78.

And 21b78 × 4 = 87b12 works fine ... because 78 × 4 = 312, giving us the '12' we need as the final two digits in the answer.

Now notice that we then carry the '3' from the 312 and add it to b × 4, to get an answer that ends in b! The only possibility for this is 9! (9 × 4 + 3 = 39).

So, the number is found: 21978.

Check it: 21978 × 4 = 87912.

So, far I have convinced myself that there are no 2- or 3-digit versions of this problem.

I still have to look at 4-digit, 6-digit and so on!

Bye for now.