1967 AMC 12 Problem 1

Attempt Problem 1 of the 1967 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1967 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

The three-digit number 2a32a3 is added to the number 326326 to give the three-digit number 5b9.5b9. If 5b95b9 is divisible by 9,9, then a+ba+b equals:
22
44
66
88
99
Answer: C
Concepts:place valuedivisibility
Small hint:
Use the digit sum of 5b95b9
Big hint:
After finding b,b, compare the tens digits in the addition
Solution:
Divisibility by 99 requires 5+b+95+b+9 to be a multiple of 9.9. Since bb is a digit, this gives b=4.b=4. The addition is then 2a3+326=549,2a3+326=549, so a=2.a=2. Hence a+b=6.a+b=6.
Therefore, the correct answer is C.
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