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22 Sep 2023, 03:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
25% (02:10) correct
75%
(02:09)
wrong
based on 83
sessions
History
Date
Time
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Not Attempted Yet
How many unordered pairs of positive integers exist such that their least common multiple is 400?
A. 9
B. 15
C. 22
D. 23
E. 45
A. 9
B. 15
C. 22
D. 23
E. 45
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22 Sep 2023, 13:24
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Bunuel
I think this is one of the trickiest questions I've solved in a long time
, but the solution is very simple and the question aims to test the very fundamental concepts.400 = 4 * 10 * 10 = 4 * 2 * 5 * 2 * 5 = \(2^4 * 5^2\)
Now by the definition of LCM we know that of the two numbers in the pair, one of the numbers must at least contain \(2^4\) in its prime factorized form. Similarly, \(5^2\) will also be present in the prime factorized form of one of the numbers.
Case 1
Let's assume that out of the two numbers in the unordered pair, one number contains both \(2^4\) & \(5^2\) in its prime factorized form. Hence, that number can be represented as \(2^4 * 5^2\). The other number can only contain powers of 2 and 5 (i.e. the other number cannot have any other prime factors except for 2 and 5). The available powers of 2 are \(2^0, 2^1, 2^2, 2^3,\) and \(2^4\), and the available powers of 5 include \(5^0, 5^1,\) and \(5^2\).
So to form the other number we can select one power of 2 from 5 available powers in \(^5C_1 = 5\) ways, and we can select one power of 5 from the three available powers in \(^3C_1 = 3\) ways.
Total number of value = \(5 * 3 = 15\)
Case 2
Out of the two numbers one number contains \(2^4\) in its prime factorized form, and the other number contains \(5^2\) in its prime factorized form.
The number that contains \(2^4 \) in its prime factorized form, can also have a 5 in it. The available powers of 5 are \(5^0\) and \(5^1\). We will not use \(5^2\) in this case, because we have handled that possibility in Case 1. So there are two ways (either \(5^0\) or \(5^1\)) in which the number can have a power of 5.
Similarly, the number that contains \(5^2 \) in its prime factorized form can have a 2 in it. The available powers of 2 are \(2^0, 2^1, 2^2,\) and \(2^3.\) Again, we will not take \(2^4\) into consideration as the possibility of \(2^4\) and \(5^2\) present in the same number was considered in Case 1. So there are four ways (\(2^0, 2^1, 2^2,\) and \(2^3.\)) in which the number can have a power of 2.
Hence, the number of possible values is 4 * 2 = 8
Total number of possible pairs = 8 + 15 = 23
Option D
Video Solution
23 Apr 2025, 00:18
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