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24 Dec 2025, 20:26
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f(100) with 100 consecutive multiples of 100 will be
100 x 2 x 100 x 3 x 100 x 4 x 100 x .... x 100 x 100
This can be written as 100^100 x (2 x 3 x 4 x .. x 100) OR 100^100 x 100!
Trailing 0s will come from combination of number of 2s and 5s which make up the zeros as 2 x 5 = 10
Similarly, 100^1 = 2^2 x 5^2, i.e. 2 zeros
100^100 = 2^200 x 5^200, which gives 200 zeros
For the remaining 100! we use the following trick:
(100 / 5^1) + (100 / 5^2) = 20 + 4 = 24 zeros
Hence, total number of zeros = 200 + 24 = 224
Answer, Option D
100 x 2 x 100 x 3 x 100 x 4 x 100 x .... x 100 x 100
This can be written as 100^100 x (2 x 3 x 4 x .. x 100) OR 100^100 x 100!
Trailing 0s will come from combination of number of 2s and 5s which make up the zeros as 2 x 5 = 10
Similarly, 100^1 = 2^2 x 5^2, i.e. 2 zeros
100^100 = 2^200 x 5^200, which gives 200 zeros
For the remaining 100! we use the following trick:
(100 / 5^1) + (100 / 5^2) = 20 + 4 = 24 zeros
Hence, total number of zeros = 200 + 24 = 224
Answer, Option D
24 Dec 2025, 20:34
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f(100) is 100*200*300*...100 times...*10000 =(100^100) * 100!
To find the number of zeroes, we know that a ten is formed by a 2 and a 5. Since 5 is the higher prime here, 5 will be the limiting factor.
We need to find out how many 5s are there in (100^100) *100!
In 100!,
[100/5] + [100/25] = 20 + 4 = 24
So there are twenty four 5s in 100!
In 100^100, we have (5^2)^100 = 200 fives.
In total we have 200+24 = 224 fives
Hence 224 zeroes exist at the end of f(100)
To find the number of zeroes, we know that a ten is formed by a 2 and a 5. Since 5 is the higher prime here, 5 will be the limiting factor.
We need to find out how many 5s are there in (100^100) *100!
In 100!,
[100/5] + [100/25] = 20 + 4 = 24
So there are twenty four 5s in 100!
In 100^100, we have (5^2)^100 = 200 fives.
In total we have 200+24 = 224 fives
Hence 224 zeroes exist at the end of f(100)
Bunuel
24 Dec 2025, 21:27
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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?
f(n) is the product of n consecutive positive multiples of n. Those multiples are n,2n,3n,4n etc
So, we can also write
f(n)= n*2n*3n*4n*..........*n.n = (n^n)*n!
Hence, f(100) = (100100)(100!)
Trailing zeros in 100100,
100= 22.52
100100 = 2200.5200
100100 will have 200 trailing zeros.
Trailing zeros in 100!,
(100/5)+(100/25)+(100/125).... = 20+4+0= 24
100! has 24 trailing zeros.
Total number of zeros in f(100) =200+24 = 224
f(n) is the product of n consecutive positive multiples of n. Those multiples are n,2n,3n,4n etc
So, we can also write
f(n)= n*2n*3n*4n*..........*n.n = (n^n)*n!
Hence, f(100) = (100100)(100!)
Trailing zeros in 100100,
100= 22.52
100100 = 2200.5200
100100 will have 200 trailing zeros.
Trailing zeros in 100!,
(100/5)+(100/25)+(100/125).... = 20+4+0= 24
100! has 24 trailing zeros.
Total number of zeros in f(100) =200+24 = 224
24 Dec 2025, 21:48
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f(n) = 1n*2n*3n*4n*...*nn = n^n*n!
f(100) = 100^100*100!
We have formula to calculate trailing zeros from 100! = 100/5+100/5^2 = 24
100^100 has 2*100 = 200 ending zeros
Total = 24+200 = 224
Answer: D
f(100) = 100^100*100!
We have formula to calculate trailing zeros from 100! = 100/5+100/5^2 = 24
100^100 has 2*100 = 200 ending zeros
Total = 24+200 = 224
Answer: D
Bunuel
