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10 Oct 2025, 00:06
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
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Question Stats:
60% (01:53) correct
40%
(01:43)
wrong
based on 199
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What is the highest integer power of of ‘2’ that can divide \((10! + 11! + 12!)\) without leaving any remainder?
(A) 8
(B) 9
(C) 12
(D) 16
(E) 25
(A) 8
(B) 9
(C) 12
(D) 16
(E) 25
10 Oct 2025, 01:05
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10! + 11! + 12! = 10!(1+11+11*12) = 10!(12 + 11*12) = 10!(144)
10! has 8 as the highest power of 2 in it
& 144 has 4 as the highest power of 2 in it
Hence, 8+4 = 12
10! has 8 as the highest power of 2 in it
& 144 has 4 as the highest power of 2 in it
Hence, 8+4 = 12
10 Oct 2025, 03:22
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Bunuel
Given that : \((10! + 11! + 12!)\)
taking 10! as common, we get
\( 10! * (1 + 11 + 11*12)\)
\( 10! * (1 + 11 + 132)\)
\( 10! * (144)\).
Then the highest power of 2 is ,
144 when divided by 2, we get 2^4.
similarly, when 10! Is successively divided by 2, we get (5+2+1)=8. That’s 2^8
Total powers of 2 = (2^4) * (2^8) = 2^12
Option C.
