Some initial terms out of city, tonight!

Number Theory Level 4

Find the number of trailing zeroes in the product $5^6 \times 6^7 \times 7^8 \times \cdots \times 49^{50} \times 50^{51}.$

423 324 285 362

3 solutions

Ayush Garg
Mar 7, 2016

Trailing zeroes will be produced by the power of 10 in the number , which in turn is 2*5. Since the number of 2's will always be greater than the the number of 5's , we only need to count the number of 5's which is the sum of powers of multiples of 5 from 5-50 where 25 and 50 have to be taken twice for the double power of 5 . Adding, we get 362.

Good solution sir.....+1!!!....u should deserve more upvotes.

rajdeep brahma - 3 years, 2 months ago

Ken Hodson
Oct 27, 2015

Certain problems, like this one, I enjoy banging out on a spreadsheet. From other problems, I've already got a listing of the prime factorizations.
For this one, I set up a column that counted each 2 in the prime factorizations of the numbers between 5 and 50. I set up another column counting each 5.
To consider the exponent, I had to take those individual counts and multiply them by their exponent.
I found 1,318 twos and 362 fives.
To answer the question, I needed to know how many 2x5 pairs existed, hence the lower number dictates.
Their are 362 pairs of 2x5, meaning their are 362 tens, meaning their are 362 zeros.

Kushagra Sahni
Oct 8, 2015

Answer should be 362 but earlier it wasn't in the options.

How u do that

Swastik Behera - 5 years, 7 months ago

Some initial terms out of city, tonight!

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