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Bunuel
The function f is defined for all positive integers n by the following rule: f(n) is the product of the distinct prime factors of n. If f(n) < 100 and n is not prime, what is the greatest possible value of f(n)?

(A) 99
(B) 95
(C) 91
(D) 87
(E) 78


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

Given: The function f is defined for all positive integers n by the following rule: f(n) is the product of the distinct prime factors of n.

Asked: If f(n) < 100 and n is not prime, what is the greatest possible value of f(n)?

(A) 99 = 3^2*11; f(99) = 3*11 = 33
(B) 95 = 5*19; f(95) = 5*19 = 95
(C) 91 = 7*13; f(91) = 7*13 = 91
(D) 87 = 3*29; f(87) = 3*29 = 87
(E) 78 = 2*39; f(78) = 2*39 = 78

IMO B
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The best way to solve this question is to backsolve from the greatest answer choice...
A 99 can be factorised as 9*11 but 9 is not prime
Move to 95=19*5 both are prime numbers
Other choices are anyway smaller than 95
Hence B

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The approach to look for is that f(n) should be greatest and should be actually the product of just prime numbers.

99 = 9 * 11 ( 9 reduces 3*3 because we're talking about prime factors.)

The number which is exclusively a product of prime numbers and greatest would be the answer.

95 = 19 * 5

Hope this helps!


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Bunuel
The function f is defined for all positive integers n by the following rule: f(n) is the product of the distinct prime factors of n. If f(n) < 100 and n is not prime, what is the greatest possible value of f(n)?

(A) 99
(B) 95
(C) 91
(D) 87
(E) 78


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

99 cannot be the answer since it's not a product of all distinct prime even if we make it mulatiple of prime =11*3*3

However 95 is an answer since 19*5 all distinct prime numbers therefore our answer
THerefore IMO B
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