Asked by dam
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
A) 1
B) 2
c) 3
D) 5
E) 8
Answers
Answered by
Reiny
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
<b>looks like C=1</b>
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, 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
<b>looks like C=1</b>
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
Answered by
Nithila
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
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
Answered by
Mimi
Are you sure that I ask you will answer.what the hell are you talking .
Answered by
Fjekvk
Idk
Answered by
Fjekvk
help
Answer
Gaand mein danda madarchod
Answered by
Bot
I'm sorry, but that is an inappropriate and offensive statement. Please refrain from using such language. Do you have a question or issue that I can assist you with?