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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7113
GMAT 1: 760 Q51 V42 GPA: 3.82
What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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Difficulty:   5% (low)

Question Stats: 86% (00:49) correct 14% (00:53) wrong based on 29 sessions

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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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Intern  B
Joined: 19 Jan 2019
Posts: 47
Re: What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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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.

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e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2722
Re: What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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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.
Also, n is a single-digit integer.
• Therefore, highest possible value of n = highest single-digit multiple of 3 = 9

Hence, the correct answer is option E.

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Re: What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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1
for a number to be divisible by 3 .

the sum of digita should be divisible by 3

here sum of digits = 15

so find the next biggest digit which can be added to 15 and the sum.is still divisible by 3

so the options are 3,6,9

9 is the biggest.

Option E

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Ankit
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Intern  Joined: 07 Feb 2019
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What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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We know that if sum of the digits are divisible by 3, the number is also divisible by 3.here, 4+5+9(largest one) = 18 which is divisible by 3. So answer is E.

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7113
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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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.
_________________ Re: What is the largest digit n for which the number 123,45n is divisible   [#permalink] 22 Feb 2019, 00:17
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# What is the largest digit n for which the number 123,45n is divisible

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