Monday Math 10

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 4e=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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