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07 Jul 2022, 01:30
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D
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Difficulty:
45%
(medium)
Question Stats:
70% (02:05) correct
30%
(02:13)
wrong
based on 153
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A is the product of the first 100 multiples of 8. How many zeros would be there at the end of A?
(A) 10
(B) 11
(C) 12
(D) 24
(E) 25
Are You Up For the Challenge: 700 Level Questions
(A) 10
(B) 11
(C) 12
(D) 24
(E) 25
Are You Up For the Challenge: 700 Level Questions
07 Jul 2022, 02:28
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A laborious process, which is also error-prone, is to list all the multiples of 8 that ends with 0 (or 00).
Alternatively, realizing that we have an oversupply of 2's, we just need to figure out how many 5's are there in the product of the first 100 multiples of 8.
8 does not contribute any 5's. We need to count the number of 5's in 1*2*3*...*100
A systematic way to do so is
Firstly, 100/5 = 20
Secondly, 100/25 = 4
(And 25*5 is already larger than 100)
So the number of 5's is 20+4 = 24.
The answer is (D).
Alternatively, realizing that we have an oversupply of 2's, we just need to figure out how many 5's are there in the product of the first 100 multiples of 8.
8 does not contribute any 5's. We need to count the number of 5's in 1*2*3*...*100
A systematic way to do so is
Firstly, 100/5 = 20
Secondly, 100/25 = 4
(And 25*5 is already larger than 100)
So the number of 5's is 20+4 = 24.
The answer is (D).
General Discussion
07 Jul 2022, 03:27
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A = 8 x 16 x 24 ...... x 800
To find the number of trailing 0(s), we need combination of 2 and 5 and since 8 already has 2 as a factor, we need to count the number of 5(s) in first 100 multiples of 8 to get the number of trailing 0(s)
We will get 5 in each 5th multiple of 8 (40,80,120...) and hence a 0 will be added at the end at each 5th multiple of 8 since its a product
So in this way we get 100/5 = 20 zeros
But, 25th (5x5), 50th (5x5x2), 75th (5x5x3) and 100th (5x5x4) multiples of 8 have 2 fives so an extra zero needs to be added to our tally so another 4
Total zeros = 24
Answer - D
Posted from my mobile device
To find the number of trailing 0(s), we need combination of 2 and 5 and since 8 already has 2 as a factor, we need to count the number of 5(s) in first 100 multiples of 8 to get the number of trailing 0(s)
We will get 5 in each 5th multiple of 8 (40,80,120...) and hence a 0 will be added at the end at each 5th multiple of 8 since its a product
So in this way we get 100/5 = 20 zeros
But, 25th (5x5), 50th (5x5x2), 75th (5x5x3) and 100th (5x5x4) multiples of 8 have 2 fives so an extra zero needs to be added to our tally so another 4
Total zeros = 24
Answer - D
Posted from my mobile device
