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If 7ab8 is a 4-digit number that is divisible by 12 and a=b+1, then what can be the value of a+b?
divisible by 12 => divisible by 3 ,4.
so, sum of the digits divisible by 3 .
since, 7+8=15 (divisible by 3 )
a+b must be divisible by 3 .
let , a+b =3 and a=b+1
so, b= 1 ( b8 i.e. 18 not divisible by 4 , rejected )
let , a+b =6 and a=b+1
so, b= fraction (rejected )
let , a+b =9 and a=b+1
so, b= 4 (48 divisible by 4 ,accepted )

a+b =9
correct answer C
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Given that 7ab8 must be divisible by 12, we know it must be divisible by both 3 and 4.

For divisibility by 4, the last two digits, "b8", must be divisible by 4. Therefore, b can only be 0, 4, or 8.

For divisibility by 3, the sum of the digits 7 + a + b + 8 must be divisible by 3. Since 7 + 8 = 15 is divisible by 3, the sum a + b must also be divisible by 3.

Since a = b + 1, we now check values for b:

b must make "b8" divisible by 4,

a + b must be divisible by 3.

When b = 4, a becomes 5 (since a = b + 1). This satisfies both conditions: 48 is divisible by 4, and 5 + 4 = 9 is divisible by 3.

Thus, a + b = 9.

Answer: C (9)
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