12 Days of Christmas GMAT Competition 2025 - Day 8: The function f(n)

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Bunuel

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24 Dec 2025, 10:00

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Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

GMAT Club Official Explanation:

Let's break down the question:

According to the problem, f(100) is the product of 100 consecutive positive multiples of 100. So:

f(100) = 100 * 200 * 300 * ... * 10,000 =

= (100 * 1) * (100 * 2) * (100 * 3) * ... * (100 * 100) =

= 100^100 * 100!

Next:

100^100 = (10^2)^100 = 10^200, which ends with 200 zeros.

100! ends with 100/5 + 100/25 = 20 + 4 = 24 zeros.

Thus, the total number of zeros at the end of f(100) = 10^200 * (something with 24 zeros at the end) is:

200 + 24 = 224 zeros.

Answer: D.

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ManifestDreamMBA

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24 Dec 2025, 09:16

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f(100) = 100 * 200 * 30 * .... * 10000
= 100^100 * 1*2*#...*100
=10^200 *100!

Number of zeroes in 100! = 100/5 + 100/25 = 20+4 = 24
Number of zeroes in 10^200 = 200
Total zeroes = 200+24 = 224

Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

Archit3110

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24 Dec 2025, 09:24

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The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

for f(100) 100^100 * ( 100)!
trailing zeros will be 100!
100/5 + 100/25 = 24
and
100^100 will have
10^200 zeros
200+24 ; 224
f(100) has 224 zeros.
OPTION D is correct

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24 Dec 2025, 09:29

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f(100)=(100*1)(100*2)...(100*100)
=100^100(100!)
100! has 24 zeros and 100^100 has 200 zeros

200+24=224
hence answer = 224

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24 Dec 2025, 09:38

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This means that f(100)= 100^100*100!
Consider a trailing zero is formed by 2*5
100^100= (2^2x5^2)^100= 2^200x5^200
This means the exponents of 5 are 200 (We use exponents of 5 because 2 tends to have more exponents)
On 100! exponents can be found through 100/5=20 plus 20/5=4= 20+4= 24 trailing zeros
Total trailing zeros= 200+24= 224
Ans D

Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

Kinshook

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24 Dec 2025, 09:40

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f(n) = n*2n*3n*4n*...nn = n^n*n!

f(100) = 100^100*100!

Number of 0's in 100^100 = (10^2)^100 = 10^200 = 200
Number of 0's in 100! = Highest power of 5 in 100! = 20 + 4 = 24

Number of 0's in f(100) = 200 + 24 = 224

IMO D

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24 Dec 2025, 09:40

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Here f(100) means 100 multiples of 100 starting from 100 x 1
Hence it is 100 x 200 x 300 ...... 10000
now taking 100^100 aside which has 200 zeros( 2 per multiple of 100) balance is 100! which has 24 zeros
Hence totally its 224 (Option D)

HarishChaitanya

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24 Dec 2025, 10:20

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that means f(n) is equals to n!

number of zero's in a n! is equal to [n/5]+[n/25]+.....i, until i =0

here f(100)= [100/5]+[100/25], which equals to 24.

Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

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Let's write out the sequence in scientific notation to keep better track of how many 10s are within f(100).

Note that what the question is asking for is how many factors of 10 are contained within f(100), because each factor of 10 will leave a trailing 0.

So f(100) = 10^2*1*10^2*2*10^2*3...10^2*100

So whenever we multiply terms with the same base, we can add the exponents. For 100 terms of 10 with the exponent 2, we will get 100*2= 10^200. (1*2*3...*100)
So, how many 10s are in the product of all the numbers from 1 to 100? Since 10 = 2*5, this depends on the limiting factor of 5 as there are plenty 2s. To find out the number of 5s within that product, we can use the formula:
100/5^1 = 20
100/5^2 = 4

So there are 24 10s.

200 + 24 = 224, answer D.

sitrem

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24 Dec 2025, 10:49

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consecutive multiples of n are: n, 2n, 3n, ...
f(n) = n * 2n * 3n * ... * (n*n) = n^n * n!

f(100) = 100^100 * 100! = 10^200 * 100!

trailing zeros come only from factors of 10 (therefore factors of 2 and 5). The smaller number between the factors of 2 and 5 is the amount of zeros. There are always more 2s than 5s so we can just count the 5s.

10^200 -> 200 zeros

100! -> how many factors of 5 are in 100! -> 100/5 + 100/25 = 24 zeros

200 + 24 = 224

answer D

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24 Dec 2025, 11:11

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difficult to understand the wordings;
f(n) = product of n consecutive positive multiples of n -> f(100) = 100*200*300*400*500*600*.....100*100
this can be re-written as f(100) = (1*2*3*4*5.....*100)*(100*100*100*.......*100) = 100!*100^100
finding the trailing zeroes for the above
100! -> 20 + 4 = 24 (no. of 5's as is the limiting factor)
100^100 = (2*5)^200 => 200 (no. of 5's and 2's both present in equal quantities)
total trailing zeroes -> 200+24 = 224

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24 Dec 2025, 11:25

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f(n)= (n*1)*(n*2)....(n*n)
f(n)=(n^n)*(1*2*...n)

f(100)=(100^100)*(1*2*...100)
f(100)=(10^200)*(1*2..100)

now 100! has 24 nos. of 5, so 24 nos. of 10.
so total =200+24=224

Ans D

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24 Dec 2025, 11:54

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Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

Testing out a value like n=5, we would get:

$f(n)=5*10*15*20*25$. If we factor out a 5 from each multiple here, we get: $f(n)=5^5(1*2*3*4*5)=5^5(5!)$

This gives us a general notation for f(n) as $f(n)=n^n(n!)$. Subbing in 100 for n, we get $f(100)=100^{100}(100!)$

Ok, lets now take each portion of that equation separately and figure out how many zeroes it contributes. A trailing zero is created every time 5x2 happens.

$100=5^2*2^2$ thus $100^{100}=5^{200}*2^{200}$

For 100! we can use the dividing trick as follows:

$\frac{100!}{5}=20$

$\frac{100!}{25}=4$

$\frac{100!}{125}=0$

Summing these, we get 24. So 100! contributes 24 zeroes and $100^{100}$ contributes 200. Together, there will be 224 zeroes, thus D is our answer.

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24 Dec 2025, 12:56

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Solving this questions requires use of the concept of finding factors within a factorial.
First translate the question: "product of n consecutive multiples of n"
This means, f(n) = (n*1)*(n*2)*(n*3)*.....(n*n) = (n^n)*(1*2*3*....*n) = (n^n)*n!
Therefore, f(100) = (100^100)*(100!).

Let's find the number of zeroes in 100!. So we'll try to find the number of 2s and 5s.
Note: Don't just try to find the number of 10s with this method. It will only give the multiples of 10 like 10,20,30, etc.
We need to check all the number. For example: 4 and 25.

Number of 2s in 100! :-
100/2 + (100/2)/2 + .... until we can't divide it anymore.
So,
50 + 25 + 12 + 6+ 3 + 1 = 97
So there are 97 2s in 100!

Similarly number of 5s in 100! :-
20 + 4 = 24.

So 100! can have a total of 24 zeroes.

Now, 100^100 = (10^2)^100 = 10^200 --> 200 zeroes.

Therefore, total number of zeroes in f(100) = 200 +24 = 224.

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24 Dec 2025, 14:10

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multiple of 100
100*1
100*2
100*3
up to 100*100

so basically its 100! multiplied by 100*100
the number of zeros in 100! are 24
every multiple of 100 has 2 zero and there are total 100 of it. 100*2=200
200+24= 224

Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

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24 Dec 2025, 15:36

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Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

trick is in defining the function
eg--- f(5)= 1x5.2x5.3x5.4x5.5x5
so f(100)= 1.100 x 2x100. .......100x100
= 100^100 x factorial(100)
we for zeroes we need to find the power of 5 in this expression
f(100)= (2^2.5^2)^100 x 100!
power of 5 = 5^200 x 5^ 24= 5^224
zeroes= 224
ans=D

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24 Dec 2025, 18:27

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If n were 5 then f(n)= product of first 5 multiples of 5.
As in, 5×10×15×20×25=5^5×5!
So, f(100)=100^100×100!
Trailing zeroes of 100! can be found by- 100/5 + 100/25=24
100^100=10^200 with 200 zeroes
Thus f(100) has total of 224 trailing zeroes

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24 Dec 2025, 18:54

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Bunuel

The function f(n) is defined for all positive integers n, where n > 1, as the product of n consecutive positive multiples of n. How many zeros does f(100) end with?

A. 24
B. 26
C. 143
D. 224
E. 2400