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Trailing zeros (Posted on 2020-10-25) Difficulty: 2 of 5
Let m and n be positive integers such that m>n and the number

22220038m - 22220038n

has eight zeros at its end. Show that n>7.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 1 of 1
We can factor this difference as 

22220038^n * (22220038^d - 1) where d = m - n > 0

For the difference to end in 8 zeros it must be a multiple of 10^8 and so must have at least 8 2's (and 8 5's) among its factors. Now, 22220038 is divisible by 2 but not by 4 (it's 11110019 * 2), so the first term contributes at most n 2's. The second term is odd and so contributes none. 

For the overall to have 8 factors of 2, then, it must be that n >= 8, or that n > 7.

  Posted by Paul on 2020-10-26 09:28:48
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