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Algebra AMC-8 Math Olympiad

Divisibility | AMC 8, 2014 |Problem 21

Try this beautiful problem from Algebra based on multiplication and divisibility of Two given numbers. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra based on multiplication and divisibility of two given numbers.

Multiplication and Divisibility- AMC 8, 2014


The  7-digit numbers 74A52B1 ana 326AB4C are each multiples of 3.which of the following could be the value of c ?

  • 1
  • 2
  • 3

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:1

AMC-8, 2014 problem 21

Challenges and Thrills of Pre College Mathematics

Try with Hints


Use the rules of Divisibility ……..

Can you now finish the problem ……….

If both numbers are divisible by 3 then the sum of their digits has to be divisible by 3……

can you finish the problem……..

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8… and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10… and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7… and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1. so the answer is 1

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