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What is the largest digit n for which the number 123,45n is

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by Max@Math Revolution » Wed Feb 20, 2019 5:20 pm
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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
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by Brent@GMATPrepNow » Thu Feb 21, 2019 11:38 am
Max@Math Revolution wrote:[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
RULE: if the SUM of the digits of a number is divisible by 3, then that number is divisible by 3
Take, for example, the number 1,001,226
1 + 0 + 0 + 1 + 2 + 2 + 6 = 12
Since 12 is divisible by 3, we know that 1,001,226 is divisible by 3

If 123,45n is divisible by 3, then 1 + 2 + 3 + 4 + 5 + n must be divisible by 3
Simplify: 15 + n must be divisible by 3
If n = 9, then the sum of the digits of 123,45n = 24, which is divisible by 3

Answer: E

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by Max@Math Revolution » Thu Feb 21, 2019 10:58 pm
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Recall that a number is divisible by 3 if and only if the sum of its digits is divisible by 3. The sum of the digits of 123,45n is 1 + 2 + 3 + 4 + 5 + n = n + 15. This is divisible by 3 exactly when n is divisible by 3.
So, n must be a multiple of 3.
Thus, the largest digit, n is 9.

Therefore, the answer is E.
Answer: E
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by Scott@TargetTestPrep » Sun Feb 24, 2019 5:30 am
Max@Math Revolution wrote:[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
Since the sum of the digits of a number divisible by 3 equal a multiple of 3, we have:

1 + 2 + 3 + 4 + 5 + n = 15 + n

Since 15 is a multiple of 3, then n also has to be a multiple of 3. Therefore, the largest digit for n is 9.

Answer: E

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