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Zeroes On My Mind

Find the smallest integer N such that N! terminates in 50 zeros.

 

Source: Charlotte Hartman...many years ago


Hint: A zero is produced by the product of what two primes?

 


Solution Commentary: Though a zero is produced by the product of a 2 and a 5, we can focus on the number of 5's (as there will be many more 2's available).

So, our new question: What is the smallest N such that N! has 50 factors of 5.

Consider [N/5] + [N/25] + [N/125] + ..., which provides a convenient way to tease out the number of factors of 5 in N!, as each of the expressions is contributing a "new" factor of 5. If you are not sure of this, let N = 125. Then, [150/5] = 30, which picks up the 5's for 5, 10, 15, 20, 25, 30,...145, 150 in 150!, while [150/25] = 6 picks up the "new" 5's for 25, 50, 75, 100, 125, 150. And, [150/125] = 1 picks up the "new" 5 for 125. Thus, in 150!, there are 30 + 6 + 1 = 37 fives. Neat, huh!

So, now play with the expression until you get the correct value of N for the stated problem.