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

[Max@Math Revolution]

[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

[Max@Math Revolution]

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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.

[deloitte247]

123, K50 is a 6-digit integer.
K can be anything between 0 and 9.
Option A: 2 can always divide 123, k50 regardless of the value of k since it ends with '50' which is even. Thus, 2 is always a factor.

Option B: 3 can divide 123, k50 regardless if the sum of all integers, (i.e 1+2+3+k+5+0) equals to a factor of 3 like 36, 12 etc. For this to occur, k= {1,4,7}. Thus, 3 can be a factor of 123, k50.

Option C: 4 can never divide 123, k50 for all values of k. This is because the possible three rear number are not divisible by 4. (i.e 150, 250, ...) Thus, 4 cannot be a factor

Option D: 5 can always divide 123, k50 and thus can be a factor.

Option E: For 6 to divide 123, k50 , the condition for divisibility by 3 (option B) must be met, and 123, k50 has to be even. This permutation exists (e.g 123450), thus 6 can be a factor.

Therefore, option C is correct

[Jeff@TargetTestPrep]

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