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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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20 Oct 2017, 08:19
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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

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Hasan Mahmud

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Joined: 02 Aug 2009
Posts: 6559
Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co  [#permalink]

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20 Oct 2017, 08:32
1
1
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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20 Oct 2017, 08:48
1
2
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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Re: Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co  [#permalink]

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30 Apr 2018, 02:31
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

This is a very tricky question.. Probably could not be solved if seen first time.

Any cube of the a^3k where a is a prime number will have 3k +1 factors.
For e.g. Factors of 3^3 = 1, 3, 3^2, 3^3
Factors of 3^6 = 1, 3, 3^2, 3^3, 3^4, 3^5, 3^6

So, we have to find the numbers of the form 3k+1 from the given numbers
7 = 3*2 +1
16 = 3*5+1
20 =/= 3k+1
22 = 3*7+1

So, p,q,s could be a perfect cube...
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Re: Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co  [#permalink]

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30 Apr 2018, 02:43
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

Another comprehensive solution could be..

Number of factor of a number a^k*b^m*c^n.......= (1+k)(1+m)(1+n)...... , where a,b,c, are prime numbers.

So, if a number has (1+k)(1+m)(1+n)...... factors then the number is of the form a^k*b^m*c^n.......

Now lets do this for the factors provided.

7 = (1+6) : So the number will be of the form a^6 ........ a perfect cube.

16 = (1+15): So the number will be of the form a^15 ........ a perfect cube.
16= (8)(2) = (1+7)(1+1): So the number will be of the form a^7*b^1
16 = (4)(4) = (1+3)(1+3): So the number will be of the form a^3 * b^3
and various other forms can be formed.....e.g. 2*2*4 , 2*2*2*2, ...

20 = 1+19 : So the number will be of the form a^19
20 = 5*4 = (1+4) (1+3): So, the number will be of the form a^4 * b^3
and various other forms can be formed ... e.g 5*2*2, 10*2
But none will give a perfect cube.

22 = 1+21: So the number will be of the form a^21............. a perfect cube.
22 = 11*2 = (1+10)(1+1): So, the number will be of the form a^10*b^1

Hence p,q and s could be a perfect cube..
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Re: Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co  [#permalink]

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01 May 2018, 09:50
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

To determine the number of total factors of a number, we add 1 to the number of each unique prime factor and multiply. Also, recall that a perfect cube has unique prime factors that are in quantities of a multiple of 3. Thus, any of these values p, q, r, or s must possess both concepts combined: i.e., the number of factors must be 1 more than a multiple of 3.

So, for instance, 2^3 is a perfect cube, and it has 3 + 1 = 4 total factors.

2^6 is a perfect cube, and it has 6 + 1 = 7 prime factors.

Thus, we see that any number that has a total number of factors that is “1 more” than a multiple of 3, is a perfect cube.

We are given that p has 7 factors; subtracting 1 from 7 gives us 6, which is a multiple of 3. Thus, p is a perfect cube.

Similarly, q has 16 factors; subtracting 1 from 16 gives us 15, which is a multiple of 3. Thus, q is a perfect cube.

We see that r is not a perfect cube because 1 less than the number of factors is 20 - 1 = 19, and 19 is not a multiple of 3.

Finally, s has 22 factors; subtracting 1 from 22 gives us 21, which is a multiple of 3. Thus, s is a perfect cube.

So p, q, and s could all be a perfect cube.

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Re: Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co &nbs [#permalink] 01 May 2018, 09:50
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