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

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

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New post 20 Feb 2019, 18:37
00:00
A
B
C
D
E

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

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New post 20 Feb 2019, 19:08
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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Re: What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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New post 20 Feb 2019, 19:33

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.

Answer: 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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New post 20 Feb 2019, 19:54
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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What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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New post 21 Feb 2019, 09:28
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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Re: What is the largest digit n for which the number 123,45n is divisible  [#permalink]

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New post 22 Feb 2019, 00:17
=>

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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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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