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24 Nov 2023, 02:30
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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:
25%
(medium)
Question Stats:
76% (01:07) correct
24%
(01:29)
wrong
based on 135
sessions
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If x = 31! – 18, then x is divisible by which of the following
I. 6
II. 9
III. 12
A. I only
B. II only
C. III only
D. I and II
E. I, II and III
I. 6
II. 9
III. 12
A. I only
B. II only
C. III only
D. I and II
E. I, II and III
24 Nov 2023, 10:17
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The answer is D.
x = 31! - 18 = 18(31*30*....*1 - 1)
Since consecutive numbers don't have any common factors, x will be divisible by all the factors of 18, but by no other numbers from 1 to 31, barring 18.
x = 31! - 18 = 18(31*30*....*1 - 1)
Since consecutive numbers don't have any common factors, x will be divisible by all the factors of 18, but by no other numbers from 1 to 31, barring 18.
24 Nov 2023, 10:48
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x = 31! – 18,
I: remainder \(\frac{(31! - 18) }{ 6}\) = 0 as both 31! and 18 are divisible by 6
II: remainder when \(\frac{(31! - 18) }{ 9}\) = 0 as both 31! and 18 are divisible by 9
III: Remainder when \(\frac{(31! - 18) }{ 12}\) = 6 as 18 is not divisible by 12
Therefore, answer is D
I: remainder \(\frac{(31! - 18) }{ 6}\) = 0 as both 31! and 18 are divisible by 6
II: remainder when \(\frac{(31! - 18) }{ 9}\) = 0 as both 31! and 18 are divisible by 9
III: Remainder when \(\frac{(31! - 18) }{ 12}\) = 6 as 18 is not divisible by 12
Therefore, answer is D
