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06 Sep 2020, 23:13
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
48% (02:29) correct
52%
(02:20)
wrong
based on 145
sessions
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If P is the product of all the even numbers between 1 and positive integer n, how many trailing zeros does P have? (The number of trailing zeros of an integer is the number of zeros at its end. For example, 360 has 1 trailing zero.)
(1) \(n^2\) < 100
(2) The product of all the integer numbers from 1 to \(n^2\) has 6 trailing zeros.
(1) \(n^2\) < 100
(2) The product of all the integer numbers from 1 to \(n^2\) has 6 trailing zeros.
06 Sep 2020, 23:55
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Statement 1: If n^2<100, n<10; product of all even numbers from 1 to 9 does not have 5, so number of trailing zeroes = 0. Sufficient.
Statement 2: (n^2)! has 6 trailing zeroes, which means n^2 must be 25,26,27,28, or 29. But n is an integer, therefore n^2 must be 25, which means n=5. Trailing zeroes in this case is 0 as well. Sufficient.
Correct option: D
Statement 2: (n^2)! has 6 trailing zeroes, which means n^2 must be 25,26,27,28, or 29. But n is an integer, therefore n^2 must be 25, which means n=5. Trailing zeroes in this case is 0 as well. Sufficient.
Correct option: D
07 Sep 2020, 20:56
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Please note, a trailing zero is the result of the product of a 2 and a 5.
Constraints: P is the product of all even numbers between 1 and n. n is positive.
Statement 1: \(n^2 < 100\)
This means that n is less than 10 and the positive integers are 2, 4, 6 and 8.
P = 2 *4 * 6 * 8 = 384
There are no trailing 0's. (Since there are no 5's, we cannot get a single 0).
Therefore Statement 1 Alone is Sufficient. Answer options could be A or D.
Statement 2: the product of 1 to \(n^2\) has 6 trailing 0's
1 to \(n^2\), means the factorial of \(n^2\)
Counting of 5's in a factorial gives us the trailing 0's. eg 5! has a 5 so, there is 1 trailing 0. Similarly 10! has 2 trailing 0's (10 has a 5 and 5 contributes to 1 more)
in this manner, 25! has 6 0's(two 5's in 25, one in 20, one in 15, one in 10, and 5 itself)
Similarly, 26!, 27!, 28! and 29! will also have 6 trailing 0's
So, \(n^2\) = 25, 2 6, 27, 28 or 29. Since n is an integer, the square root of 25 is the only possibility.
Therefor n = 5. The even integers from 1 to 5 are 2 and 4. p = 2 * 4 = 8. Again no trailing zeroes.
Therefore Statement 2 Alone is Sufficient.
Option D
Arun Kumar
Constraints: P is the product of all even numbers between 1 and n. n is positive.
Statement 1: \(n^2 < 100\)
This means that n is less than 10 and the positive integers are 2, 4, 6 and 8.
P = 2 *4 * 6 * 8 = 384
There are no trailing 0's. (Since there are no 5's, we cannot get a single 0).
Therefore Statement 1 Alone is Sufficient. Answer options could be A or D.
Statement 2: the product of 1 to \(n^2\) has 6 trailing 0's
1 to \(n^2\), means the factorial of \(n^2\)
Counting of 5's in a factorial gives us the trailing 0's. eg 5! has a 5 so, there is 1 trailing 0. Similarly 10! has 2 trailing 0's (10 has a 5 and 5 contributes to 1 more)
in this manner, 25! has 6 0's(two 5's in 25, one in 20, one in 15, one in 10, and 5 itself)
Similarly, 26!, 27!, 28! and 29! will also have 6 trailing 0's
So, \(n^2\) = 25, 2 6, 27, 28 or 29. Since n is an integer, the square root of 25 is the only possibility.
Therefor n = 5. The even integers from 1 to 5 are 2 and 4. p = 2 * 4 = 8. Again no trailing zeroes.
Therefore Statement 2 Alone is Sufficient.
Option D
Arun Kumar
