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dave13
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How many trailing Zeroes does 49! + 50! have? 07 Aug 2018, 11:09
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Bunuel
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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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Bunuel why 51 won't contribute to the number of zeros at the end of the number ? :? i dont get


\(\frac{49}{5} = 9\)

\(\frac{51}{5^2}\) = 2

\(9+2 =11\) :?
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How many trailing Zeroes does 49! + 50! have? 08 Aug 2018, 00:51
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dave13
Bunuel
chetan2u
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.


Bunuel why 51 won't contribute to the number of zeros at the end of the number ? :? i dont get


\(\frac{49}{5} = 9\)

\(\frac{51}{5^2}\) = 2

\(9+2 =11\) :?
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The formula works only when we have a factorial !
when we have an integer , it has to be factorised.

consider 6!
6!= 6*5*4*3*2*1
no of zeroes = no of 5*2 pairs
we have only one in this case , so no of trailing zeroes = 1
using the formula
6/5=1

now consider 6
6=2*3
doesn't have a five in it , and therefore doesn't have a trailing zero

similarly 51 is an integer
factors of 51 are 1,3,17,51
no 5's or 2's , so no trailing zeroes
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How many trailing Zeroes does 49! + 50! have? 08 Aug 2018, 01:18
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CounterSniper many thanks for great explanation :) i have one question if after factoring out we have 49!(1+50) = 49! *51

why Bunuel uses 49! here twice ---> Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\) we have only one 49! left after factoring out :?

have a good day :)
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How many trailing Zeroes does 49! + 50! have? 08 Aug 2018, 01:24
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dave13
CounterSniper many thanks for great explanation :) i have one question if after factoring out we have 49!(1+50) = 49! *51

why Bunuel uses 49! here twice ---> Trailing zeros in 49!: \(\frac{49}{5}+\frac{49}{5^2}=9+1=10\) we have only one 49! left after factoring out :?

have a good day :)
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Thats how the formula works .

you keep on dividing the numerator with increasing powers 5 .

you might find this useful !!

https://gmatclub.com/forum/gmat-math-book-87417.html
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How many trailing Zeroes does 49! + 50! have? 30 Apr 2025, 08:10
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