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The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i

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Math Revolution GMAT Instructor
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The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i  [#permalink]

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New post 02 Apr 2018, 02:16
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[GMAT math practice question]

The number 123,k50 is a 6-digit integer, and k is a positive 1 digit integer. Which of the following cannot be a factor of 123,k50?

A. 2
B. 3
C. 4
D. 5
E. 6

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The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i  [#permalink]

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New post 02 Apr 2018, 02:22
MathRevolution wrote:
[GMAT math practice question]

The number 123,k50 is a 6-digit integer, and k is a positive 1 digit integer. Which of the following cannot be a factor of 123,k50?

A. 2
B. 3
C. 4
D. 5
E. 6


As we're asked 'which cannot be a factor' we'll look to the relevant divisibility tests.
This is a Precise approach.

The number ends is 0 and therefore will always be divisible by 2 and 5. (A), (D) are eliminated.
If the sum of 1,2,3,k,5,0 is divisible by 3, then the number will be divisible by 3. This happens for at least one k, for example k = 1. (B) is eliminated.
If the number is divisible by both 2 and 3, it is divisible by 6. Based on the above, we know that this can happen and therefore (E) is eliminated.

(C) must be our answer.

Alternatively, a number is divisible by 4 if its last two digits are divisible by 4. Since 50 is not divisible by 4, neither is our number.
Once again, (C) is our answer.
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Re: The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i  [#permalink]

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New post 02 Apr 2018, 02:55
MathRevolution wrote:
[GMAT math practice question]

The number 123,k50 is a 6-digit integer, and k is a positive 1 digit integer. Which of the following cannot be a factor of 123,k50?

A. 2
B. 3
C. 4
D. 5
E. 6


A number is divisible by 4 only if its last two digits are divisible by 4. Since 50 is not divisible by 4.
(C) is our answer
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Re: The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i  [#permalink]

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New post 04 Apr 2018, 02:51
=>

The last two digits tell us whether the number is divisible by \(4\).
Since \(50\) is not a multiple of \(4\), the number cannot be a multiple of \(4\).

Therefore, the answer is C.

Answer: C

Let’s see why the number could be divisible by each of the other options:

A: Since the units digit is an even number, the whole number is a multiple of \(2\).
B: A number is divisible by \(3\) if the sum of its digits is divisible by \(3\). If \(k = 4\), then the sum of the digits is \(1 + 2 + 3 + 4 + 5 + 0 = 15\), which is a multiple of \(3\), and so the number is a multiple of \(3\).
D: Since the units digit is a multiple of \(5\), the number is a multiple of \(5\).
E: If \(k = 4\), the number is divisible by \(3\) as seen in part B. Since it is also divisible by \(2\) (see part A), the number is divisible by \(6.\)
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Re: The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i  [#permalink]

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New post 17 Apr 2018, 16:30
MathRevolution wrote:

The number 123,k50 is a 6-digit integer, and k is a positive 1 digit integer. Which of the following cannot be a factor of 123,k50?

A. 2
B. 3
C. 4
D. 5
E. 6


We may recall that an integer is divisible by 4 if the last two digits are divisible by 4.

For example, 1,224 is divisible by 4 since 24/4 = 6.

Since 50 are the last two digits of 123,k50, and 50 is not divisible by 4, 123,k50 is not divisible by 4 regardless what digit k is.

Answer: C
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Re: The number 123,k50 is a 6-digit integer, and k is a positive 1 digit i &nbs [#permalink] 17 Apr 2018, 16:30
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