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05 Aug 2021, 05:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:18) correct
27%
(02:14)
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based on 73
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What is the least five-digit integer which when divided by 63, 56 and 42 leaves remainder 1 in each case?
(A) 10082
(B) 10081
(C) 10071
(D) 10007
(E) 10001
(A) 10082
(B) 10081
(C) 10071
(D) 10007
(E) 10001
05 Aug 2021, 06:05
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Smallest five-digit number = 10000
Let us divide this by the L.C.M of 63, 56, 42
=> 63 = 3^2 * 7
=> 56 = 2^3 * 7
=> 42 = 2 * 3 * 7
LCM = 2^3 * 3^2 * 7 = 504
Dividing 10000 by 504 leaves a remainder 424, Subtract 424 from 10000
=> 10000-424 = 9576
So this means 9576 is divisible by 504.
9576+504 = 10080 (which will also be divisible by 504, 63, 56, and 42)
To find the number which leaves reminder 1, we have to add 1 to 10080.
So the answer is 10081.
OA should be B
Let us divide this by the L.C.M of 63, 56, 42
=> 63 = 3^2 * 7
=> 56 = 2^3 * 7
=> 42 = 2 * 3 * 7
LCM = 2^3 * 3^2 * 7 = 504
Dividing 10000 by 504 leaves a remainder 424, Subtract 424 from 10000
=> 10000-424 = 9576
So this means 9576 is divisible by 504.
9576+504 = 10080 (which will also be divisible by 504, 63, 56, and 42)
To find the number which leaves reminder 1, we have to add 1 to 10080.
So the answer is 10081.
OA should be B
30 Jan 2022, 21:27
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If 1 is deducted from the correct answer, the number should be divisible by 63, 56 and 42 without leaving a remainder. Because the number (upon deduction of 1) is divisible by 63, the number should also be divisible by 9 (which is a factor of 63). Amongst the answer choices, only choice B is divisible by 9 upon deduction of 1. Hence, the answer is B.
