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17 Nov 2022, 01:34
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D
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Difficulty:
95%
(hard)
Question Stats:
48% (03:13) correct
52%
(02:54)
wrong
based on 85
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For any integer k, k> 1, the "width" of k is the number of primes (not necessarily distinct) whose product is k. For example, the width of 24 is 4 since 24 = 2x2x2 x 3. If p and q are positive integers and p + 5q < 2000, what is the maximum possible sum of the width of p and the width of q?
(A) 5
(B) 6
(C) 15
(D) 17
(E) 18
(A) 5
(B) 6
(C) 15
(D) 17
(E) 18
27 Nov 2022, 20:41
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First of all, we have to understand the meaning of the "width" which is nothing but number of prime factors of any particular number, no matter if the factors are the same.
Then we have to maximize [p+5q] in such a way that it remains below 2000.
Note that the smallest prime is 2, so to maximize the width we will use no other prime than 2.
The first step is to maximize the "p" because if we maximize "q" then "5q" will not give us the maximum possible width in a way that [p+5q] remains below 2000.
Maximizing "p": \(2^{10}\) gives us 1024, but we can't go beyond that as it will cross the 2000 mark.
So, the maximum width of "p" is 10.
Then, we have to maximize "q" to the extent that [1024+5q] remains below 2000.
\(2^7\) gives us 128 and we can't go beyond that as [1024+5q] will result in a number greater than 2000.
So, the maximum width of "q" is 7.
Thus the maximum possible sum of the width of p and the width of q = 10 + 7 = 17
Option D is the right answer.
Then we have to maximize [p+5q] in such a way that it remains below 2000.
Note that the smallest prime is 2, so to maximize the width we will use no other prime than 2.
The first step is to maximize the "p" because if we maximize "q" then "5q" will not give us the maximum possible width in a way that [p+5q] remains below 2000.
Maximizing "p": \(2^{10}\) gives us 1024, but we can't go beyond that as it will cross the 2000 mark.
So, the maximum width of "p" is 10.
Then, we have to maximize "q" to the extent that [1024+5q] remains below 2000.
\(2^7\) gives us 128 and we can't go beyond that as [1024+5q] will result in a number greater than 2000.
So, the maximum width of "q" is 7.
Thus the maximum possible sum of the width of p and the width of q = 10 + 7 = 17
Option D is the right answer.
General Discussion
14 Feb 2024, 09:15
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