SOLUTION: Let n! Equal n×(n-1) ×…×2×1. For example, 4! = 4×3×2×1 = 24. A clerk performs the computation: 1% of 2% of 3% of … 100% of n and gets the number 99! How many

Algebra ->  Probability-and-statistics -> SOLUTION: Let n! Equal n×(n-1) ×…×2×1. For example, 4! = 4×3×2×1 = 24. A clerk performs the computation: 1% of 2% of 3% of … 100% of n and gets the number 99! How many      Log On


   

Question 1170704: Let n! Equal n×(n-1) ×…×2×1. For example, 4! = 4×3×2×1 = 24.

A clerk performs the computation: 1% of 2% of 3% of … 100% of n and gets the number 99!

How many digits in n are zero?
A) 98 b) 99 c) 198 d) 199 e) 200
Answer by math_tutor2020(3817) About Me  (Show Source):
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1% of 2% of 3% of ... 100% of n = 99!

0.01*0.02*0.03*...*0.99*1.00*n = 99!

(1/100)*(2/100)*(3/100)*...*(99/100)*(100/100)*n = 99!

(1*2*3*...*99*100)/(100^100)*n = 99!

(99!*100)/(100^100)*n = 99!

100/(100^100)*n = 1 ...... note the 99! terms cancel

100*n = 1*100^100

100*n = 100^100

n = (100^100)/(100)

n = (100^100)/(100^1)

n = 100^(100-1)

n = 100^99

n = (10^2)^99

n = 10^(2*99)

n = 10^198

There are 198 zeros in the number n.
n is equal to 1 followed by 198 zeros.

The exponent over 10 tells us how many zeros there are.

Consider a smaller example such as 10^3 = 1000 which has three zeros.
Or consider 10^5 = 100,000 which has five zeros
The number of zeros in 10^k is k.

Answer: C) 198