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15 Jun 2020, 03:49
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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:
25%
(medium)
Question Stats:
65% (00:54) correct
35%
(01:26)
wrong
based on 99
sessions
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Not Attempted Yet
[GMAT math practice question]
\(P\) is the product of positive integers \(1\) through \(30\). How many consecutive \(0’s\) are there in the units digit of \(P\)?
A. \(4\)
B. \(5\)
C. \(6\)
D. \(7\)
E. \(8\)
\(P\) is the product of positive integers \(1\) through \(30\). How many consecutive \(0’s\) are there in the units digit of \(P\)?
A. \(4\)
B. \(5\)
C. \(6\)
D. \(7\)
E. \(8\)
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15 Jun 2020, 14:27
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MathRevolution
Us this formula and it's an easy one:
Number of trailing zeros in n! = (n/5)+(n/5^2)+...+(n/5^k) such that n>5^k and then round down.
In this case, #of trailing zeros = (30/5)+(30/25)=6+1.2=7.2.
After rounding down, we get 7 ==> answer D.
17 Jun 2020, 01:59
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Let \(P\) be the product of integers \(1\) through \(30.\)
Then the prime factorization of \(P\) is \(P = 2^a3^b5^c7^d·…·29.\)
Define [\(x\)] as the greatest integer less than or equal to \(x\).
Then the number of the prime factors of \(2\) is \(26 = 15 + 7 + 3 + 1, \)
since \([\frac{30}{2}] = 15, [\frac{30}{2^2}] = 7, [\frac{30}{2^3}] = 3\) and \([\frac{30}{2^4}] = 1.\)
The number of prime factors of \(5\) is \(7 = 6 + 1,\)
since \([\frac{30}{5}] = 6\) and \([\frac{30}{5^2}] = 1.\)
Then we have:
\(P = 2^{26}·3b·5^7·7d·…·29 = 2^{7+19}·3b·5^7·7d·…·29 =\)
\(2^72^{19}·3b·5^7·7d·…·29 = 2^7·5^7·2^{19}·3b·7d·…·29 =10^7·2^{19}·3b·7d·…·29\), and \(P\) has \(7\) consecutive zeros in the units digit.
Since we have \(10 = 2*5\) and more \(2’s\) than \(5’s\), we can count the number of \(5’s\) only.
Therefore, D is the answer.
Answer: D
Let \(P\) be the product of integers \(1\) through \(30.\)
Then the prime factorization of \(P\) is \(P = 2^a3^b5^c7^d·…·29.\)
Define [\(x\)] as the greatest integer less than or equal to \(x\).
Then the number of the prime factors of \(2\) is \(26 = 15 + 7 + 3 + 1, \)
since \([\frac{30}{2}] = 15, [\frac{30}{2^2}] = 7, [\frac{30}{2^3}] = 3\) and \([\frac{30}{2^4}] = 1.\)
The number of prime factors of \(5\) is \(7 = 6 + 1,\)
since \([\frac{30}{5}] = 6\) and \([\frac{30}{5^2}] = 1.\)
Then we have:
\(P = 2^{26}·3b·5^7·7d·…·29 = 2^{7+19}·3b·5^7·7d·…·29 =\)
\(2^72^{19}·3b·5^7·7d·…·29 = 2^7·5^7·2^{19}·3b·7d·…·29 =10^7·2^{19}·3b·7d·…·29\), and \(P\) has \(7\) consecutive zeros in the units digit.
Since we have \(10 = 2*5\) and more \(2’s\) than \(5’s\), we can count the number of \(5’s\) only.
Therefore, D is the answer.
Answer: D
