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15 Apr 2019, 01:20
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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:
68% (01:33) correct
32%
(01:41)
wrong
based on 190
sessions
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Not Attempted Yet
[GMAT math practice question]
What is the largest multiple of 24 that can be written using each of the digits 0, 1, 2, … , 9 exactly once?
A. 9876543210
B. 9876543120
C. 987654102
D. 1234567890
E. 1234567980
What is the largest multiple of 24 that can be written using each of the digits 0, 1, 2, … , 9 exactly once?
A. 9876543210
B. 9876543120
C. 987654102
D. 1234567890
E. 1234567980
15 Apr 2019, 01:52
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24 is 8*3
First do the divisibilty test of 8 which is check the last 3 digits if its divisible by 8
Only B survives
Then do the divisibilty test of 3. Add all the digits and check if its divisible by 3
Hence B
Posted from my mobile device
First do the divisibilty test of 8 which is check the last 3 digits if its divisible by 8
Only B survives
Then do the divisibilty test of 3. Add all the digits and check if its divisible by 3
Hence B
Posted from my mobile device
17 Apr 2019, 01:23
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=>
A number is a multiple of \(24\) precisely when it is a multiple of both \(3\) and \(8\).
The sum of \(0, 1, 2, … , 9, “0 + 1 + 2 + … + 9 = 45”\) is a multiple of \(3\). Thus, a number which uses each of the digits \(0, 1, 2, … , 9\) exactly once is a multiple of \(3\).
In order for a number to be a multiple of \(8\), the last three digits of the number must be a multiple of \(8\). And the smallest multiple of \(8\) that can be formed using three of the digits \(0, 1, 2, … , 9\) is \(120\).
Thus, the biggest possible multiple of \(24\) is \(9876543120\).
Therefore, B is the answer.
Answer: B
A number is a multiple of \(24\) precisely when it is a multiple of both \(3\) and \(8\).
The sum of \(0, 1, 2, … , 9, “0 + 1 + 2 + … + 9 = 45”\) is a multiple of \(3\). Thus, a number which uses each of the digits \(0, 1, 2, … , 9\) exactly once is a multiple of \(3\).
In order for a number to be a multiple of \(8\), the last three digits of the number must be a multiple of \(8\). And the smallest multiple of \(8\) that can be formed using three of the digits \(0, 1, 2, … , 9\) is \(120\).
Thus, the biggest possible multiple of \(24\) is \(9876543120\).
Therefore, B is the answer.
Answer: B
