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05 Aug 2021, 02:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
63% (02:30) correct
37%
(02:16)
wrong
based on 150
sessions
History
Date
Time
Result
Not Attempted Yet
What is the greatest four-digit positive integer which when divided by 10, 15, 21 and 28 leaves 4, 9, 15 and 22 as remainders, respectively ?
(A) 9654
(B) 9664
(C) 9666
(D) 9672
(E) 9864
(A) 9654
(B) 9664
(C) 9666
(D) 9672
(E) 9864
05 Aug 2021, 02:42
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The greatest 4 digits positive integer is = 9999
=> Divisor - Reminder
10 - 4 = 6
15 - 9 = 6
21 - 15 = 6
28 - 22 = 6
=> LCM (10, 15, 21, 28) = 420
420 is the least positive integer divisible by 10, 15, 21, and 28
=> Reminder when 9999 is divided by 420 = 339
=> So the greatest 4 digits positive integer divisible by 10, 15, 21, 28 = 9999-339 = 9660
Since the difference between divisor and remainder is the same, the greatest 4 digits positive integer satisfying the condition in question will be = 9660-6 = 9654
So the OA should be A
=> Divisor - Reminder
10 - 4 = 6
15 - 9 = 6
21 - 15 = 6
28 - 22 = 6
=> LCM (10, 15, 21, 28) = 420
420 is the least positive integer divisible by 10, 15, 21, and 28
=> Reminder when 9999 is divided by 420 = 339
=> So the greatest 4 digits positive integer divisible by 10, 15, 21, 28 = 9999-339 = 9660
Since the difference between divisor and remainder is the same, the greatest 4 digits positive integer satisfying the condition in question will be = 9660-6 = 9654
So the OA should be A
07 Aug 2021, 07:40
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Bunuel
Add 6 to this 4-digit positive integer. Then we know the number would be a multiple of 10, 15, 21, and 28.
The GCF of 10 = 2*5, 15 = 3*5, 21 = 3*8, and 28=4*7 is 4*3*5*7 = 420. The highest multiple of 420 under 10k is 8400 + 1260 = 9660.
Then our number is 9660 - 6 = 9654.
Ans: A
