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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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20 Oct 2017, 07:19
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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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Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co
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20 Oct 2017, 07:32
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
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20 Oct 2017, 07:48
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 Efor 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
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30 Apr 2018, 01:31
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... Answer E.
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Re: Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co
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30 Apr 2018, 01:43
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
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01 May 2018, 08:50
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. Answer: E
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Re: Numbers p, q, r and s have 7, 16, 20 and 22 factors. Which of these co &nbs
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