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Bunuel
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If n is a positive integer such that 2n has 28 positive divisors and 3 28 Dec 2021, 08:45
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32% (03:00) correct 68% (02:06) wrong based on 59 sessions

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If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?

(A) 32

(B) 34

(C) 35

(D) 36

(E) 38


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If n is a positive integer such that 2n has 28 positive divisors and 3 26 Jan 2022, 16:40
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alternate method?
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If n is a positive integer such that 2n has 28 positive divisors and 3 27 Jan 2022, 02:16
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chetan2u, yashikaaggarwal.

Would love to hear your take on this question.
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If n is a positive integer such that 2n has 28 positive divisors and 3 27 Jan 2022, 07:18
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Bunuel
If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?

(A) 32

(B) 34

(C) 35

(D) 36

(E) 38


Are You Up For the Challenge: 700 Level Questions
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A very quick method that uses logic

2n has 28 factors, so surely something in n with the extra 2 or without extra 2 gets us 7.
Why? 28=7*4=7*2*2. The number 2n could be \(a^6*b^3\) or \(a^6*b*c\), where a could be 2

If say a was something else, then we should have had 3n having number of factors multiple of 7.
Say a was 3, then a would go to power of 7 in 3n, that is \(a^7*….\), so number of factors would be a multiple of 7+1 or 8. But that is not the case.

So 2n contains \(2^6*…\).
Thus, 3*2n would not change \(2^6\), and the number of factors of 3*2n or 6n would be a multiple of 7.
Only 35 is a multiple of 7.


\(Algebraic way\)

Let the number n be \(2^{a-1}*3^{b-1}*5^c*x^d\)

Thus, number of factors of 2n or \(2^{a-1+1}*3^{b-1}*5^{c-1}*x^{d-1}\) = \((a+1)bcd=28\)………(i)
Similarly, the number of factors of 3n or \(2^{a-1}*3^{b}*5^{c-1}*x^{d-1}\) = \((b+1)acd=30\)…..(ii)

As all are positive numbers divide i by ii => \(\frac{(a+1)b}{a(b+1)}=\frac{28}{30}=\frac{7*4}{6*5}\)
So, a=6 and b=4.
Or, \((a+1)b*30=(b+1)a*28……..30ab+30b=28ab+28a………2ab+30b=28a…….ab+15b=14a……a(14-b)=15b\)
Put b as few values and you will find only a=6 and b=4 will make sensible pair.


Another method
Also shown above.

First analyse 2n having 28 factors and 3n having 30 factors.
1) If 2 or 3 is not included in n, then a new factor should increase the total number of factors by at least 2 times as number of factors will get multiplied by (1+1)
2) Since 2 increases by one and 3 decreases by one in 2n as compared to 3n, let us look for the correct pair in 28 and 30. n is \(2^a*3^b*x^c\), then
2n or \(2^{a+1}*3^b*x^c\) has 28 factors =7*4=7*2*2
3n or \(2^{a+1}*3^b*x^c\) has 30 factors =6*5=2*3*5=15*2=10*3
We can see that 7*4 and 6*5 match the observation above.
7*4=(6+1)(3+1) and 6*5=(5+1)(4+1)
So, the number is \(2^5*3^3\) and 6n is \(2^6*3^4\), and number of factors are 7*5=35
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If n is a positive integer such that 2n has 28 positive divisors and 3 28 Jan 2022, 08:36
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chetan2u
Bunuel
If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?

(A) 32

(B) 34

(C) 35

(D) 36

(E) 38


Are You Up For the Challenge: 700 Level Questions

A very quick method that uses logic

2n has 28 factors, so surely something in n with the extra 2 or without extra 2 gets us 7.
Why? 28=7*4=7*2*2. The number 2n could be \(a^6*b^3\) or \(a^6*b*c\), where a could be 2

If say a was something else, then we should have had 3n having number of factors multiple of 7.
Say a was 3, then a would go to power of 7 in 3n, that is \(a^7*….\), so number of factors would be a multiple of 7+1 or 8. But that is not the case.

So 2n contains \(2^6*…\).
Thus, 3*2n would not change \(2^6\), and the number of factors of 3*2n or 6n would be a multiple of 7.
Only 35 is a multiple of 7.


\(Algebraic way\)

Let the number n be \(2^{a-1}*3^{b-1}*5^c*x^d\)

Thus, number of factors of 2n or \(2^{a-1+1}*3^{b-1}*5^{c-1}*x^{d-1}\) = \((a+1)bcd=28\)………(i)
Similarly, the number of factors of 3n or \(2^{a-1}*3^{b}*5^{c-1}*x^{d-1}\) = \((b+1)acd=30\)…..(ii)

As all are positive numbers divide i by ii => \(\frac{(a+1)b}{a(b+1)}=\frac{28}{30}=\frac{7*4}{6*5}\)
So, a=6 and b=4.
Or, \((a+1)b*30=(b+1)a*28……..30ab+30b=28ab+28a………2ab+30b=28a…….ab+15b=14a……a(14-b)=15b\)
Put b as few values and you will find only a=6 and b=4 will make sensible pair.


Another method
Also shown above.

First analyse 2n having 28 factors and 3n having 30 factors.
1) If 2 or 3 is not included in n, then a new factor should increase the total number of factors by at least 2 times as number of factors will get multiplied by (1+1)
2) Since 2 increases by one and 3 decreases by one in 2n as compared to 3n, let us look for the correct pair in 28 and 30. n is \(2^a*3^b*x^c\), then
2n or \(2^{a+1}*3^b*x^c\) has 28 factors =7*4=7*2*2
3n or \(2^{a+1}*3^b*x^c\) has 30 factors =6*5=2*3*5=15*2=10*3
We can see that 7*4 and 6*5 match the observation above.
7*4=(6+1)(3+1) and 6*5=(5+1)(4+1)
So, the number is \(2^5*3^3\) and 6n is \(2^6*3^4\), and number of factors are 7*5=35
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That's clear, thank you!
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If n is a positive integer such that 2n has 28 positive divisors and 3 29 Mar 2026, 07:31
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