If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (2024)

If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (1)

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If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (2)

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If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q,[#permalink]Updated on: 16 Jun 2021, 10:16

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If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (14)

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If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (15)If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (16)If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (17)75%(hard)

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43%(01:53)correctIf N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (18)57%(01:53) wrongIf N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (19)based on83sessions

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If \(N = 11^{(p + 7)}7^{(q – 2)}5^{(r + 1)}3^s\) is a perfect cube, where p, q, r and s are positive integers, then the smallest value of p + q + r + s is :

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

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Originally posted by hD13 on 16 Jun 2021, 08:36.
Last edited by Bunuel on 16 Jun 2021, 10:16, edited 8 times in total.

Renamed the topic and edited the question.

If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (20)

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chetan2u

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Re: If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q,[#permalink]16 Jun 2021, 09:55

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hD13 wrote:

If \( N=11^{p+7}*7^{q–2}*5^{r+1}*3^s \)is a perfect cube, where p,q,r and s are positive integers,
then the smallest value of p+q+r+s is :

a) 5
b) 6
c) 7
d) 8
e) 9...

Since N is in its prime factorisation form, each prime factor should have a power that is a multiple of 3..
1) p+7 will be 9, when p=2
2) q-2 will be 0, when q=2
3) r+1 will be 3, when r = 2
4) s will be 3, when s=3, as s is positive.

\(p+q+r+s=2+2+2+3=9\)

E
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1. Absolute modulus:. 2. Fractions: 3. Percentage increase: 4. Combinations:

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If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (23)

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Re: If N = 11^(p + 7)7^(q 2)5^(r + 1)3^s is a perfect cube, where p, q,[#permalink]06 Aug 2022, 11:10

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If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (25)

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Re: If N = 11^(p + 7)7^(q 2)5^(r + 1)3^s is a perfect cube, where p, q,[#permalink]

06 Aug 2022, 11:10

If N = 11^(p + 7)7^(q – 2)5^(r + 1)3^s is a perfect cube, where p, q, (2024)
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