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13 Dec 2016, 10:52
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C
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Difficulty:
45%
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Question Stats:
63% (01:23) correct
37%
(01:29)
wrong
based on 522
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What is the greatest value of q such that 9^q is a factor of 21! ?
A. 1
B. 3
C. 4
D. 5
E. 6
A. 1
B. 3
C. 4
D. 5
E. 6
13 Dec 2016, 13:15
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\(9 = 3^2\)
First we need to find max power of \(3\) in \(21!\)
\([\frac{21}{3}] + [\frac{21}{3^2}] = 7 + 2 = 9\)
But we are asked about max power of 9 not 3, so we need to divide our power of 3 by 2 and take integer value.
\([\frac{9}{2}] = 4\)
\(q=4\)
Answer C.
First we need to find max power of \(3\) in \(21!\)
\([\frac{21}{3}] + [\frac{21}{3^2}] = 7 + 2 = 9\)
But we are asked about max power of 9 not 3, so we need to divide our power of 3 by 2 and take integer value.
\([\frac{9}{2}] = 4\)
\(q=4\)
Answer C.
General Discussion
13 Dec 2016, 14:08
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First find the number of 3s in 21! when prime factoring each term. E.g. how many powers of 3 in 21!.
21 -3*7
20 - NA
19 - NA
18 - 2*(3^2)
17 - NA
16 - NA
15 - 5*3
14 - NA
13 - NA
12 - (2^2)*3
11 - NA
10 - NA
9 - 3^2
8 - NA
7 - NA
6 - 2*3
5- NA
4 - NA
3 - 3^1
2 - NA
1 - NA
Totalling up the number of 3s gives 3^9 = (3^2)^4.5 = 9^4.5. Therefore the highest power of 9 must be 4 (5 is larger than 4.5).
21 -3*7
20 - NA
19 - NA
18 - 2*(3^2)
17 - NA
16 - NA
15 - 5*3
14 - NA
13 - NA
12 - (2^2)*3
11 - NA
10 - NA
9 - 3^2
8 - NA
7 - NA
6 - 2*3
5- NA
4 - NA
3 - 3^1
2 - NA
1 - NA
Totalling up the number of 3s gives 3^9 = (3^2)^4.5 = 9^4.5. Therefore the highest power of 9 must be 4 (5 is larger than 4.5).
