A number has seven digits. The product of the first three digits is equal to the product of the last three digits is equal to the product of the middle three digits. The digits are distinct. What is the middle (fourth) digit?

  • $\begingroup$ I've posted my solution below but you're welcome to share your own. No doubt someone will find a better and quicker argument. $\endgroup$ Feb 19 '15 at 21:45

Using the letters a..g for the digits we have abc = cde = efg. Observe each digit features in at most two products: c and e are shared between two products, the rest of the digits appear in only one product.

(Note there will be multiple solutions—if any—because given one solution reversing the number or swapping a and b will yield another)

Recalling the digits are distinct, conclude 0 cannot be one of them, because it would zero one or two but not all of the products. Similarly 5 and 7 can't be digits, by considering prime factors. One but not all the products would be divisible by 5 or 7.

The digits are thus 1234689 in some order. Let w be the product abc = cde = efg. By thinking about prime factors again, we conclude 8 divides w and 9 divides w. Thus 72 divides w. Looking at the powers of 2 and 3 in the smaller digits, conclude w is exactly 72.

With these digits 72 can be expressed as only three products 1*8*9, 3*4*6 and 2*4*9. The shared digits (c and e) are thus 4 and 9. The digit in between and the middle digit overall is 2.

Indeed 1892436 is a solution.

  • 4
    $\begingroup$ You didn't even give us a chance to solve it! $\endgroup$ Feb 19 '15 at 17:14
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    $\begingroup$ This just seems to me a way of earning reputation!....... very disappointed!! :/ $\endgroup$ Feb 19 '15 at 17:27
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    $\begingroup$ So downvote the answer and the question $\endgroup$ Feb 19 '15 at 17:44
  • $\begingroup$ @TheDragonista it seems he has done this twice already $\endgroup$
    – Taylor C
    Feb 19 '15 at 20:31
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    $\begingroup$ I'm here because I enjoy sharing and solving interesting puzzles. It doesn't matter who answers them first. I shared my solution because I think it's a neat argument. If you want a competition, you should enter a competition. $\endgroup$ Feb 19 '15 at 22:01

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