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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co

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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co [#permalink]

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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these could be a perfect cube?

A. p and q
B. q and r
C. r and s
D. p, q and r
E. p, q and s
[Reveal] Spoiler: OA

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Hasan Mahmud

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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co [#permalink]

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Mahmud6 wrote:
Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these could be a perfect cube?

A. p and q
B. q and r
C. r and s
D. p, q and r
E. p, q and s



Hi...

since the Q is asking COULD, we have to find the possibilities of matching the perfect cube..
7 factors.. 1*7...... so if a number has 6 of a kind that is \(a^6\), factors = \(1+6=7...a^6= (a^2)^3\)....YES
16 factors.. 1*16...... so if a number has 15 of a kind that is \(a^{15}\), factors = \(1+15=16...a^{15}= (a^5)^3\)....YES
22 factors.. 1*21...... so if a number has 21 of a kind that is \(a^{21}\), factors = \(1+21=22...a^{21}= (a^7)^3\)....YES

p, q and s

E
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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co [#permalink]

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Mahmud6 wrote:
Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these could be a perfect cube?

A. p and q
B. q and r
C. r and s
D. p, q and r
E. p, q and s


any cube is of the form \(a^{3k}\), and hence the number of factors for this number will be \(3k+1\)

This implies that the number of factors when divided by \(3\) will leave \(1\) as remainder.

so for our question \(7\), \(16\) & \(22\) will leave a remainder of \(1\) when divided by \(3\). Hence \(p\), \(q\) & \(s\) can be a perfect cube

Option E

for the sake of explanation, \(s\) has \(22\) factors, so \(s\) can be of the form \(p_1^{21}\), this can be written as \((p_1^7)^3\) i.e a perfect cube

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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co   [#permalink] 20 Oct 2017, 08:48
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