The 7-digit numbers 74A52B1 and 326AB4C are each multiples of 3. Which of the following could be the value of C ?
A) 1
B) 2
c) 3
D) 5
E) 8
to be a multiple of 3, the sum of the digits must be a multiple of 3
For 74A52B1, 7+4+A+5+2+B+1 must divide evenly by 3
19 + A+B must divide by 3
then 18 +A+B+1 must divide by 3
A+B+1 must divide by 3
So A+B could be 2, 5, 8, 11, 14, 17, no more
in the second: 326AB4C the same rule applies
3+2+6+A+B+4+C divides by 3
15 + (A+B)+C divides by 3
A+B+C divides by 3
if A+B = 2 , then C=1
if A+B= 5 , then C=1 ***
if A+B = 8 , then C=1
looks like C=1
let's test ***
if A+B=5,
A=1, B=4, C=1
your two numbers are 7415241 and 3261441 both are divisible by 3
For 74A52B1,
Divisible by 3,
7+4+A+5+2+B+1=19+A+B……………[1]
For 326AB4C,
Divisible by 3,
3+2+6+A+B+4+C=15+A+B+C………[2]
A+B=-19…………[1]
A+B=-C-15………[2]
So,
C+15=19
C=19-15
C=4