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Four Digit Trial (Posted on 2014-06-05) Difficulty: 3 of 5
N is a 4-digit positive integer such that the sum of the four digits of N equals the product of the first two digits of N and also equals the product of the last two digits of N.

Find all N's and prove there are no others.

No Solution Yet Submitted by K Sengupta    
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Solution Non-computer solution | Comment 2 of 3 |

We're looking for a, b, c, d such that:

a + b + c + d = a * b = c * d

None of them can be 0 (if one of them was 0, they'd all have to be 0, which contradicts the condition that abcd is a 4-digit positive integer).

None of them can be 1 (if, for instance, a = 1 then 1 + b + c + d > 1 * b).

Since ordering within (a,b) and (c,d), respectively, doesn't matter, let's let a <= b and c <= d.  Assume that a != c.  Then we'd need distinct pairs of integers in the range [2,9] with the same product.  We can see by associating prime factors that the only possibilities are:

4 * 3 = 6 * 2
4 * 4 = 8 * 2
6 * 3 = 9 * 2
6 * 4 = 8 * 3
6 * 6 = 9 * 4

In none of those cases do the four digits sum to their product.  Thus a = c and b = d.

So now we have 2*a + 2*b = a*b.

If a = b, then we have 4*a = a*a, so a = 4 and the solution is 4444.  4 is an upper bound on a since we assumed a < b and if they are both > 4 then the product will exceed twice their sum. 

If a = 3, then we have 6 + 2*b = 3*b so b = 6, and solutions are 3636, 3663, 6336, and 6363.

If a = 2 then we have 4 + 2*b = 2*b so there are no additional valid solutions. 

 


 


  Posted by tomarken on 2014-06-05 17:19:25
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