"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 2

*abc*3 or 2*abc*8? (using*a, b*and*c*for for the missing digits)Now it's pretty obvious that 2

*abc*3 × 4 cannot equal 3*cba*2. So our number must be 2*abc*8.Now 2

*abc*8 × 4 = 8*cba*2 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:

20

*bc*8 × 4 = 8*cb*02 21

*bc*8 × 4 = 8*cb*12 22

*bc*8 × 4 = 8*cb*22It'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 21

*bc*8, and that 21*bc*8 × 4 has to equal 8*cb*12.To get that 1 in the answer 8

*cb*12, then*c*× 4 must end in an 8; hence*c*is either 2 or 7.So, I now have two possibilites:

21

*b*18 × 4 = 81*b*12 21

*b*78 × 4 = 87*b*12.If you start multiplying 21

*b*18 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 21

*b*78.And 21

*b*78 × 4 = 87*b*12 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.

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