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02 Dec 2022, 02:30
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B
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Difficulty:
75%
(hard)
Question Stats:
58% (03:00) correct
42%
(02:33)
wrong
based on 118
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If a, b and c are integers such that 0 < a < b < c < 2a, what is the greatest common factor of \(84^a\), \(126^b\), and \(98^c\) ?
A. \((2^a)(7^a)\)
B. \((2^b)(7^a)\)
C. \((2^b)(7^c)\)
D. \((2^a)(3^a)(7^a)\)
E. \((2^b)(3^b)(7^a)\)
A. \((2^a)(7^a)\)
B. \((2^b)(7^a)\)
C. \((2^b)(7^c)\)
D. \((2^a)(3^a)(7^a)\)
E. \((2^b)(3^b)(7^a)\)
02 Dec 2022, 02:54
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Bunuel
\(84^a=2^{2a}*3^a*7^a\)
\(126^b=2^{b}*3^{2b}*7^b\)
\(98^c=2^{c}*7^{2c}\)
Lowest factors of 2, 3 and 7 in each
2: b<c<2a, so lowest factor is \(2^b\)
3: 0<a<2b, so lowest factor is \(3^0\) or 1
7: a<b<2c, so lowest factor is \(7^a\)
Answer: \(2^b*7^a\)
You could also use small values, say 0<3<4<5<6, that is a=3, b=4 and c=5.
B
02 Dec 2022, 02:56
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Bunuel
\(84^a = (2^2 * 7 * 3)^a = (2^{2a}*7^a*3^a)\)
\(126^b = (2 * 7 * 3^2)^b = (2^b * 7^b * 3^{2b})\)
\(98^c = (2 * 7^2)^c = (2^c * 7^{2c})\)
Only powers of 2 and 7 are common
Greatest common factor = \(2^b * 7^a\)
Option B
