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23 Feb 2021, 01:15
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Difficulty:
45%
(medium)
Question Stats:
64% (01:52) correct
36%
(01:48)
wrong
based on 118
sessions
History
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Time
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Not Attempted Yet
If t is an integer, is 3^t a factor of 21! ?
(1) t is the product of two distinct single-digit prime numbers that are smaller than 7.
(2) 0 < t < 9
(1) t is the product of two distinct single-digit prime numbers that are smaller than 7.
(2) 0 < t < 9
24 Feb 2021, 01:30
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Bunuel
If t is an integer, is 3^t a factor of 21!?
The power of 3 in factorization of 21! is 21/3 + 21/3^2 = 7 + 2 = 9. (Check here: https://gmatclub.com/forum/everything-ab ... 85592.html)
(1) t is the product of two distinct single-digit prime numbers that are smaller than 7. If t = 2*3 = 6 < 9, then 3^t will be a factor of 21! but if t = 3*5 = 15 > 9, then 3^t won't be a factor of 21!. Not sufficient.
(2) 0 < t < 9. For any nonnegative t less than or equal to 9, 3^t is a factor of 21!. Sufficient.
Answer: B.
Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.
General Discussion
Updated on: 02 Mar 2025, 11:48
Originally posted by GMATinsight on 23 Feb 2021, 01:49.
Last edited by Bunuel on 02 Mar 2025, 11:48, edited 3 times in total.
Last edited by Bunuel on 02 Mar 2025, 11:48, edited 3 times in total.
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Bunuel
Question: Is \(3^t\) a factor of 21!?
Power of 3 in 21! \(= [\frac{21}{3}]+[\frac{21}{3^2}]+[\frac{21}{3^3}] = 7+2+0 = 9\)
here \([\frac{21}{3}]\) means the Greatest integer less than or equal to \(\frac{21}{3}\)
i.e. \(3^9\) is the largest power of 3 which is a factor of 21!
here \([\frac{21}{3}]\) means the Greatest integer less than or equal to \(\frac{21}{3}\)
i.e. \(3^9\) is the largest power of 3 which is a factor of 21!
Question REPHRASED: Is t ≤9?
Statement 1:t is the product of two distinct single-digit prime numbers that are smaller than 7.
i.e. t could be 2*3 = 5 OR
t could be 3*5 = 15
NOT SUFFICIENT
Statement 2: 0 < t < 9
SUFFICIENT
Answer: Option B
