How many trailing Zeroes does 49! + 50! have?

Bunuel

We calculate trailing zeroes using 5 and not 2. In this, do we assume that the number of two's will be at least equal to the 5's?

How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22


Self made

How many trailing Zeroes does 49! + 50! have?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 22

Self made

Factor out 49! from the expression: \(49! + 50!=49!(1+50)=49!*51\).

51 won't contribute to the number of zeros at the end of the number, therefore all zeros will come from 49!.

Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\)

Answer: B.

THEORY:

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.

For example 125,000 has 3 trailing zeros;

The number of trailing zeros n!, the factorial of a non-negative integer \(n\), can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\)

It's easier if we consider an example:

How many zeros are in the end (after which no other digits follow) of 32!?

\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\).

So there are 7 zeros in the end of 32!.

Another example, how many trailing zeros does 125! have?

\(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\),

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

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