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20 Feb 2019, 18:37
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
84% (00:40) correct
16%
(00:42)
wrong
based on 89
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[GMAT math practice question]
What is the largest digit \(n\) for which the number \(123,45n\) is divisible by \(3\)?
\(A. 3\)
\(B. 5\)
\(C. 6\)
\(D. 7\)
\(E. 9\)
What is the largest digit \(n\) for which the number \(123,45n\) is divisible by \(3\)?
\(A. 3\)
\(B. 5\)
\(C. 6\)
\(D. 7\)
\(E. 9\)
20 Feb 2019, 19:08
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A number is divisible by 3 if the sum of its digits is divisible by 3.
In the above question.
1+2+3+4+5+n should be a number that is divisible by 3...
And looking into the answer choices, the largest number that can take the value of n is 9, Because 1+2+3+4+5+9= 24 which is divisible by 3.
Option E is the right answer.
Posted from my mobile device
In the above question.
1+2+3+4+5+n should be a number that is divisible by 3...
And looking into the answer choices, the largest number that can take the value of n is 9, Because 1+2+3+4+5+9= 24 which is divisible by 3.
Option E is the right answer.
Posted from my mobile device
20 Feb 2019, 19:33
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Solution
Given:
- • The number 123,45n is divisible by 3
• Also, n is representing a digit
To find:
- • The largest possible value of n
Approach and Working:
We know that if a number is divisible by 3, then the sum of all the digits of the number should also be a multiple of 3.
For the number 123,45n
- • Sum of the digits = (1 + 2 + 3 + 4 + 5 + n) = 15 + n
We know that (15 + n) is a multiple of 3.
- • 15 itself is a multiple of 3.
• Hence, n should also be a multiple of 3.
- • Therefore, highest possible value of n = highest single-digit multiple of 3 = 9
Hence, the correct answer is option E.
Answer: E
