Here’s a simple problem:

Find, if possible, a five digit number such that four times that number has the same digits in reverse order.

Representing the five digits with the letters `a`, `b`, `c`, `d`, `e`, we have:

` a b c d e`

__ × 4__

` e d c b a`

First, we note that in multiplication by 4, the largest we can have as carry over to the next digit is 3 (as 4×9=36).

Now, we see that as there is no carry over into a sixth digit, `a` can only be 1 or two. And as the product of 4 and `e` must be even, we have that `a`=2, and so:

` 2 b c d e`

__ × 4__

` e d c b 2`

The fact that 4`e`=2 mod 10 means `e` is either 3 or 8. Similarly, we see in the last digit, 2×4+(carry over)=`e`, which constrains `e` to either 8 or 9. Thus `e`=8, and we obtain:

` 2 b c d 8`

__ × 4__

` 8 d c b 2`

We see now that 4×`b` has no carry over, thus meaning `b` is 0, 1, or 2. We have a carry over of 3 from the ones digit to the tens digit, and so 4×`d`+3=`b` mod 10. As 4×`d` is even, 4×`d`+3 is odd, meaning `b` is odd, and is thus 1. Therefore, we get:

` 2 1 c d 8`

__ × 4__

` 8 d c 1 2`

We now have 4×`d`+3=1 mod 10, meaning `d` is 2 or 7. We also have from the fourth digit `d`=4+(carry over). Thus, we see `d`=7, and the carry over from the third digit to the fourth is 3.

` 2 1 c 7 8`

__ × 4__

` 8 7 c 1 2`

The carry over from the tens to the hundreds is 3, and so we have 4×`c`+3=30+`c`, which has solution `c`=9, and thus our five-digit number is 21978, and 21978×4=87912.

How about if we have a six digit number, instead of five? Seven digits? If you investigate, and think about the above solution, the answer should be clear.

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