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Variety of divisors (Posted on 2015-12-03) Difficulty: 2 of 5
Find the smallest positive integer having divisors ending in 0, 1, 2, 3, ... 8, 9.

No Solution Yet Submitted by Ady TZIDON    
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Solution solution | Comment 1 of 3

The number must be divisible by 10, dealing with 1,2,5,and 0. The most economical way to churn through divisors ending in odd numbers is to use powers of 3. 2*3*5 gives us 3 and 6. 2*3^2*5 gives us 8 and 9. 2*3^3*5 supplies 4 and 7.

So the answer is 270.


  Posted by broll on 2015-12-03 11:14:52
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