Question

test: nystce multi - subject: teachers of childhood (grade 1–grade 6) part two: mathematics - pearson evaluation system
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i - subject: teachers of childhood (grade 1–grade 6) part two: mathematics
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the numeral 1a372 has a factor of 6. if a represents the digit in the thousands place, what is the sum of all possible values of a?
a. 7
b. 13
c. 15
d. 18
correct response: c. (competency 0001) for 6 to be a factor of 1a372, 1a372 must be divisible by both 2 and 3. the numeral is even, so it is divisible by 2. for the numeral to be divisible by 3, the sum of its digits must be a multiple of 3. that is, 1 + a + 3 + 7 + 2, or a + 13, must be a multiple of 3. since a is a single digit, it must be a number from 0 to 9, which means that a + 13 must be a multiple of 3 between 0 + 13 and 9 + 13. multiples of 3 between 13 and 22 are 15, 18, and 21. solving for a + 13 = 15, a + 13 = 18, and a + 13 = 21, shows that a = 2, 5, or 8. thus, the sum of all possible values of a is 2 + 5 + 8, or 15.

Explanation:

Step1: Recall divisibility rules for 6

A number is divisible by 6 if it is divisible by both 2 and 3. The number \(1A372\) is even (ends with 2), so it is divisible by 2. Now check divisibility by 3: the sum of digits must be a multiple of 3. The sum of digits is \(1 + A + 3 + 7 + 2 = A + 13\).

Step2: Find possible values of \(A\)

\(A\) is a single digit (0 - 9), so \(A + 13\) ranges from \(13\) (when \(A = 0\)) to \(22\) (when \(A = 9\)). Multiples of 3 in this range: 15, 18, 21.

• For \(A + 13 = 15\), solve \(A = 15 - 13 = 2\).
• For \(A + 13 = 18\), solve \(A = 18 - 13 = 5\).
• For \(A + 13 = 21\), solve \(A = 21 - 13 = 8\).

Step3: Sum the possible values of \(A\)

Sum the values \(2 + 5 + 8 = 15\).

Answer:

C. 15