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03 Jul 2025, 02:15
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If n has the prime factorization n=p1^a1*p2^a2⋯pk^ak
then the number of positive factors of n is (a1+1)(a2+1)⋯(ak+1)
For n to have 24 factors if (a1+1) alone = 24 then a1=23 so a number could be 2^23 or 3^23 or any prime number ^ 23 & have 24 factors in place. This is the least possible extent of the breakdown for n. So the lowest value for distinct prime factors of n is 1.
24 can also be written as 2*2*2*3 which is the maximum possible extent of its breakdown. Which means that 3 different prime numbers can have a power of 1 & another prime number can have its power raised to 2. For example, the number could be 2*3*5*7^2.
The number of factors here would be (1+1)*(1+1)*(1+1)*(2+1) = 24. Hence the number of distinct prime numbers here are 4.
The range of distinct prime numbers would hence be 4 (max)-1 (lowest) = 3 which is option (C)
then the number of positive factors of n is (a1+1)(a2+1)⋯(ak+1)
For n to have 24 factors if (a1+1) alone = 24 then a1=23 so a number could be 2^23 or 3^23 or any prime number ^ 23 & have 24 factors in place. This is the least possible extent of the breakdown for n. So the lowest value for distinct prime factors of n is 1.
24 can also be written as 2*2*2*3 which is the maximum possible extent of its breakdown. Which means that 3 different prime numbers can have a power of 1 & another prime number can have its power raised to 2. For example, the number could be 2*3*5*7^2.
The number of factors here would be (1+1)*(1+1)*(1+1)*(2+1) = 24. Hence the number of distinct prime numbers here are 4.
The range of distinct prime numbers would hence be 4 (max)-1 (lowest) = 3 which is option (C)
03 Jul 2025, 02:47
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24 factors -> (a1+1)(a2+1)..(a_k+1) = 24.
For minimum k, we can have just 1 prime : a1 + 1 = 24. So,in this case, k = 1
For maximum k, we can break 24 as
2 x 2 x 2 x 3, so 4 primes with exponents 1,1,1,2
-> k= 4
Range : 4 − 1 = 3
Answer: C
For minimum k, we can have just 1 prime : a1 + 1 = 24. So,in this case, k = 1
For maximum k, we can break 24 as
2 x 2 x 2 x 3, so 4 primes with exponents 1,1,1,2
-> k= 4
Range : 4 − 1 = 3
Answer: C
03 Jul 2025, 02:58
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if \(n = a^p*b^q*c^r*...\) (\(a,b,c,...\)are disctinct primes) then the number of positive factors of n =(p+1)(q+1)(c+1)...=24
24=23+1 \(\to\) \(n=a^{23}\) (1 prime)
24=2*12 \(\to\) \(n=a^1*b^{11}\) (2 primes)
24= 2*2*6 \(\to\) \(n=a^1*b^1*c^5\) (3 primes)
24= 2*2*2*3 \(\to\) \(n=a^1*b^1*c^1*d^2\) (4 primes - maximum possible)
\(\implies\) range = 4-1=3
Answer: C
24=23+1 \(\to\) \(n=a^{23}\) (1 prime)
24=2*12 \(\to\) \(n=a^1*b^{11}\) (2 primes)
24= 2*2*6 \(\to\) \(n=a^1*b^1*c^5\) (3 primes)
24= 2*2*2*3 \(\to\) \(n=a^1*b^1*c^1*d^2\) (4 primes - maximum possible)
\(\implies\) range = 4-1=3
Answer: C
