\( 2n = 2^p *a^q * b^r*.......*3^z\); Number of factors = (p+1)*(q+1)*......*(z+1)
\(3n = 2^{p-1} *a^q * b^r*.......*3^{z+1}\); Number of factors = (p)*(q+1)*......*(z+2)
In 3n, the exponent of 2 is reduced by 1 and exponent of 3 is increased by 1.
So i factor 28 and 30 in all possible ways and choose the combination in which one factor is reduced by 1 and other is increased by 1.
28= 4*7; 30= 5*6
4 +1 = 5 and 7-1 =6
Other way to solve the question28=1*28, 2*14, 4*7 or 2*2*7
Case 1- 28=1*28
\(2n = a^{27}\), a is prime number; a must be equal to 2, as 2n has only 1 prime factor.
\(3n = 3*2^{26}\). Number of factors = 2*27 = 54 (
Rejected)
Case 2- 28=2*14
\(2n = a*b^{13}\). Either a= 2 or b=2
1) \(2n = 2*b^{13}\); \(3n = 3*b^{13}\)
number of factors = 28 , if b is not equal to 3 (
Rejected)
number of factors =
15 , if b is equal to 3 (
Rejected)
2) \(2n = a*2^{13}\); \(3n = 3*a*2^{12}\)
number of factors = 2*2*13 =
52, if a is not equal to 3 (
Rejected)
number of factors = 3*13 =
39, if a is equal to 3 (
Rejected)
Case 3- 28=4*7
\(2n = a^3*b^6\). Either a= 2 or b=2
1) \(2n = 2^3*b^6\); \(3n = 3*2^2*b^6 \)
number of factors = 2*3*7 =
42 , if b is not equal to 3 (
Rejected)
number of factors = 3*8=
24 , if b is equal to 3 (
Rejected)
2) \(2n = a^3*2^6\); \(3n = 3*a^3*2^5 \)
number of factors = 2*4*6 =
48 , if a is not equal to 3 (
Rejected)
number of factors = 5*6=
30, if a is equal to 3 (
What we looking for)abcdddddd
nick1816
Comparing to 2n, in 3n there is decrease in the exponent of 2 by 1 and increase in the the exponent of 3 by 1
28=1*14, 2*14,4*7 or 2*2*7
30= 1*30, 2*15, 3*10,5*6 or 5*3*2
Hence n should be equals to 2^5 *3^3
6n = 2^6 * 3^4
Numbers of factor of 6n= 7*5=35
Hi
nick1816,
Could you explain the solution in more detail? Didn't quite get the method you used. Thanks!