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

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firefox300

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25 Dec 2025, 05:33

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f(n=100)=100*200*300*...*10000 = 100^100 * (1*2*3*...*100) = 100^100 * 100!

Zeros in 100^100: 2*100=200
Zeros in 100!: 100/5 + 100/(5^2) = 20 + 4 = 24

Total number of zeros: 24+200=224

The correct answer is D

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25 Dec 2025, 05:38

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as per given conditions f(n)=1n*2n*3n*.............(n-1)n*(n*n) product of n consecutive n multiples of n
so f(100)=100*200*300.............(99*100)*(100*100)
=(1*2*3......99*100)*100^100
=factorial (100)*100^100=100!*10^200
no of zeroes in 100!=[100/5]+[100/25]=24 where [] is greatest integer function
total no of trailing zeroes=24+200=224
so D

remdelectus

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25 Dec 2025, 05:59

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f(n)=H^nk=1(n.k)=n^n.n!
f(100)=100^100x100!
100=2^2x5^2. 100^100
V2=200, V5=200
100!
V5(100!)=(100/5)+(100/25)=20+4=24
V2(100!)=50+25+12+6+3+1=97
V2=200+97=297
V5=200+24=224

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25 Dec 2025, 06:02

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We’re asked for the number of trailing zeros in f(100).

Defining the Function f(100)
The problem defines f(n) as the product of n consecutive positive multiples of n.
For n = 100: The multiples of 100 are: 100, 200, 300, ...
The product of the first 100 multiples is: f(100) = 100 x 200 x 300 x ....x (100 x 100)
We can factor out 100 from each of the 100 terms: f(100) = 100^(100) x (1 x 2 x 3 x .... x 100)
f(100) = (10^2)^(100) x 100!
f(100) = 10^(200) x 100!

Counting the Zeros.
The number of trailing zeros in a number is determined by the number of times it is divisible by 10, which is equivalent to the number of pairs of factors of 2 and 5.
Since factors of 5 are rarer than factors of 2, we count the total number of factors of 5.
=>Zeros from 10^{200} This term clearly contributes exactly 200 zeros.
=>Zeros from 100! Similarly dividing it until we get < 5 as quotient and adding them through Legendre's Formula to find the exponent of the prime p=5 in n!
= (100/5) + (100/25) + (100/125) = 20 + 4 + 0 = 24
So, 100! ends in 24 zeros.

Now, To find the total number of zeros, we add the zeros from both parts of the product= 200 + 24 = 224

Answer should be 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

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Rishm

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25 Dec 2025, 06:03

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f(100) is basically 100^100*100! and 100^100 is 5^200, then divide 100/5 and divide 100/25 and then u will get 24 and add it with 200 and u get 224 trailing zeroes

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25 Dec 2025, 06:15

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f(n) = n*2n*3n...(n*n) = n^n(1*2*3*n) = n^n*n!

f(100) = 100^100*100!

To find the number of zeros in a number, we need to know the power of 2's and 5's of that number. Since we need both 2's ans 5's and since power of 5's<power of 2's, just need to know the power of 5's.

100^100 = (2^2*5^2)^100 = 5^200

100! = 100/5 = 20/5 = 4 => 20+4 = 24

Total power of 5s = Total zeros = 200+24 = 224 (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

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msignatius

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25 Dec 2025, 07:12

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Okay, I used what I would like to call the "cheesed this" method on this, lol. I hope it would help anyone as stuck with concepts as I can be at times, as I was with this one.

We know the function states for the value we pick, we will multiply that value by as many consecutive multiples.

So, for f(100), that's essentially 100 multiplied by 100 consecutive multiples of 100. Or 100, 200, 300, 400... etc.

Now, as we start from 100, we'll go all the way up to 10,000.

For each value from 100 to 900, two extra zeroes will be added.

That's 2*9 = 18 zeroes.

And for each value from 1,000 to 9,900, for these 90 values, 3 digits each will be added for the increments of 1,000 = 1,000, 2000 .... 9,000 (that's 3*9 = 27 zeroes).

And for all the numbers in between, like 1,100, 1,600 etc - that's 81 of the remaining 90 - we will have 2 digits each added.

That's 2*81 = 162 zeroes.

Then, for the final one, we add 4 zeroes (10,000).

That's a total of 18 + 27 + 162 + 4 = 211.

Now, we don't have that in the answer choices. But logic dictates, that the number is going to be nearest to this (211 -> 224), as the options are too far apart. We can go with this, and, thus, cheesing it, get it right

- option D.

Now, here's my guess on where the remaining 13 zeroes went.

We know, multiplying a 400 with a 2500 = 100,000. Now, that doesn't just add 4 zeroes, it adds 5. Like this, there should be 12 more combinations that'll get you the answer.

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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sanjitscorps18

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25 Dec 2025, 07:39

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Given f(n) = 1.n * 2.n * 3.n *...... * n.n

f(100) = 1.100 * 2.100 * ......... * 100*100
= 100^100 * (1*2*3.......*100) = 100^100 * 100!

For 100!, we can figure out how many 5s are present in 100! and it comes to 24

100^100 = 2^200 * 5^200, hence 5s would be 200 for 100^100

Hence in total the number of 5s in the product can be expressed as 5^200 * 5^24 = 5^224

We can figure out the trailing zeroes with 5s as 2s would be greater than 5s in 100! and for the other part they are same in number.

Hence the number would have 224 trailing zeroes.

Option 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

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raffaeleprio

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25 Dec 2025, 07:47

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Let's write the multiplication of the multiples of 100 as:
f(100) = 100^100 * 100! ( as we multiply 100 times the number 100 times all the numbers from 1 to 100)

100^100 = 10^200 = 200 zeros
100! = ( we take for every tens a 0 and a 2*5 so total 2 zeros) and we get 24

so total 224 zeros.

IMO D!

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25 Dec 2025, 07:56

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We need to find the trailing zeroes in $f(100)=(1*100) * (2*100)*(3*100)*.....*(99*100)*(100*100).$

Well, first we have 100 of 100's. This means we get $2*100=200$ zeroes.

Now let's look at the zeroes we obtain from our coefficients, namely $1/2/3/.../98/99/100$. To get a zero at the end, we need to obtain a 10, which is made by a 2 and a 5. Since there're way fewer fives than twos, we'll count the fives. There're 20 fives from 1 to 100, and there's one additional in each full square multiple (i.e., $25=5*5,$ and also 50, 75 and 100). This means we get an additional $20+4=24$ zeroes.

Finally, $200+24=224$ zeroes, and the answer is D.

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25 Dec 2025, 08:15

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so the solution is:

we need to count number of zeros in

100*200*300*400.......................10000

Now we will take 100 common

100^100(1*2*3*4.............100)

now in 1*2*3*4.... we can check there will be 24 zeros so
we can write as

100^100(10^24)
100^100(100^12)

100^112

therefore total zeros would be 2*112 = 224

Hence (D) answer

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04 Feb 2026, 15:00

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Hi Bunuel,

Shouldn't it be explicitly mentioned that consecutive multiples of n starting from n itself ?

Bunuel

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HNS

Hi Bunuel,

Shouldn't it be explicitly mentioned that consecutive multiples of n starting from n itself ?

No, it’s fine as it is. “n consecutive positive multiples of n” is read as n, 2n, ..., n^2. For example, 100 consecutive positive multiples of 100 are 100, 200, ..., 10,000 because 100 is the first positive multiple of 100. Why mention this explicitly?