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Question Stats:
67% (02:12) correct
33%
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If 48^x is a factor of 200! and x is a positive integer, what is the maximum possible value of x?
A. 197
B. 97
C. 95
D. 49
E. 8
The OA will be automatically revealed on Tuesday 18th of November 2025 11:05:22 PM Pacific Time Zone
A. 197
B. 97
C. 95
D. 49
E. 8
The OA will be automatically revealed on Tuesday 18th of November 2025 11:05:22 PM Pacific Time Zone
17 Nov 2025, 12:43
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200! has 3^97 and 2^197
But since we need four 2s for 48 to be able to divide -> the limiting power will be that of 2.
197/4 = 49
Hence, 48^49 will be able to divide 200!
But since we need four 2s for 48 to be able to divide -> the limiting power will be that of 2.
197/4 = 49
Hence, 48^49 will be able to divide 200!
18 Nov 2025, 02:13
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ExpertsGlobal5
48^x can be factorized into (2^4x) * (3^x)
If 200! Is successively divided by 3 , we get
= 66+ 22 + 7 + 2
= 97
But, if 200! Is successively divided by 2, we get
= 100 + 50 + 25 + 12 + 6 + 3 + 1
= 197
This value 197 is for 2. But we need (2^4)
2^ 197 = 2^ (49*4) . 2^1
= 2^4 (49) . 2^1
The maximum value of x can be 49.
Option D
