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02 Jul 2025, 10:52
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We have to find the smallest and largest possible distinct prime factors that n can have.
If \(n=p_1^{a_1}p_2^{a_2}p_3^{a_3}\dots p_k^{a_k}\)
then, number of factors = \((a_1+1) (a_2+1) (a_3+1)\dots (a_k+1) = 24\)
Now, the minimum number of factors happens when all 24 of them come from one exponent:
\(24=(23+1)\)\(\implies{n=p^{23}}\) \(\to\)only 1 prime factor
For the maximum number of prime factors, we break 24 into the smallest numbers possible:
\(24=2\times2 \times 2 \times 3\)
\(\implies a_1+1=2, a_2+1=2, a_3+1=2, a_4+1=4\)
\(\implies a_1=1, a_2=1, a_3=1, a_4=2\) \(\to\)4 prime factors
\(\therefore \) Range = 4-1 = 3 (C)
If \(n=p_1^{a_1}p_2^{a_2}p_3^{a_3}\dots p_k^{a_k}\)
then, number of factors = \((a_1+1) (a_2+1) (a_3+1)\dots (a_k+1) = 24\)
Now, the minimum number of factors happens when all 24 of them come from one exponent:
\(24=(23+1)\)\(\implies{n=p^{23}}\) \(\to\)only 1 prime factor
For the maximum number of prime factors, we break 24 into the smallest numbers possible:
\(24=2\times2 \times 2 \times 3\)
\(\implies a_1+1=2, a_2+1=2, a_3+1=2, a_4+1=4\)
\(\implies a_1=1, a_2=1, a_3=1, a_4=2\) \(\to\)4 prime factors
\(\therefore \) Range = 4-1 = 3 (C)
02 Jul 2025, 10:54
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First, we recognize that,
Any positive integer, N, can be represented as a multiple of it's prime factors to some power
\(N = P^a*Q^b*R^c\), where P,Q,R are prime numbers, and a,b,c are their respective powers
\(# of factors of N = (a+1)*(b+1)*(c+1)\)
Now, we are given that "n" has 24 positive factors and we need to find the RANGE of the # of prime factors that "n" can have,
We can write the 24 factors of "n", as \(P^2^3\), and we will get 24 positive factors -----> Note in this case we only have 1 prime factor, P
To find the maximum number of prime factors, we need to split 24 into the maximum number of its factors,
It can be written as, \(24 = 2*2*2*3\),
which means "n" can be written as \(P^1*Q^1*R^1*S^2\) ------> In this case, we have 4 different prime factors, P,Q,R, and S
So to caluclate the range of the # of prime factors, we have 4 - 1 = 3
So, the answer is C.
Any positive integer, N, can be represented as a multiple of it's prime factors to some power
\(N = P^a*Q^b*R^c\), where P,Q,R are prime numbers, and a,b,c are their respective powers
\(# of factors of N = (a+1)*(b+1)*(c+1)\)
Now, we are given that "n" has 24 positive factors and we need to find the RANGE of the # of prime factors that "n" can have,
We can write the 24 factors of "n", as \(P^2^3\), and we will get 24 positive factors -----> Note in this case we only have 1 prime factor, P
To find the maximum number of prime factors, we need to split 24 into the maximum number of its factors,
It can be written as, \(24 = 2*2*2*3\),
which means "n" can be written as \(P^1*Q^1*R^1*S^2\) ------> In this case, we have 4 different prime factors, P,Q,R, and S
So to caluclate the range of the # of prime factors, we have 4 - 1 = 3
So, the answer is C.
02 Jul 2025, 11:33
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Bunuel
Let the positive integer n has 24 positive factor. Let p,q,r,s are all prime factors.
P^23 = 24 factors. This has only one prime factors.
P^1 * Q^11 = 2*12 = 24 factors.
P^2 * Q^7 = 3*8 = 24 factors.
P^3 * Q^5 = 4*6 = 24 factors.
P^1 * Q^1 * R^1 * S^2 = 2*2*2*3 = 24 factors. There are 4 prime factors.
Range of prime factors = 4 - 1 = 3 factors.
Option C
