swerve wrote:The sum of the digits of a three-digit number is 11. What is the product of the three digits?

1) The number is divisible by 5.
2) The hundreds digit is twice the tens digit.

The OA is C.

Please, can someone explain this question? I need help to understand why C is the correct answer. Thanks.

Given: The sum of the digits of a three-digit number is 11.

We have to find out the product of the three digits.

Let's take each statement one by one.

1) The number is divisible by 5.

A number is divisible by 5 if its units digit is 0 or 5. If the units digit is 0, the product of the three digits; however, if it 5, the product of the three digits may or may not be 0. Example: 650 => product = 0; 425 => product = 40. No unique answer. Insufficient.

2) The hundreds digit is twice the tens digit.

There can be a couple of examples such as 425 => product = 40; 218 => product = 16. No unique answer. Insufficient.

(1) and (2) together

Say the units digit is 0, then the sum of the hundreds digit and the tens digit = 11. From (2), we know that the hundreds digit is twice the tens digit. Say tens digit = x, thus, the hundreds digit = 2x.

=> x + 2x = 3x = 11 => x = 11/3 = Not an integer, which is not possible. Thus, the units digit cannot be 0.

Thus, the units digit is 5. Thus, the sum of the hundreds digit and the tens digit = 11 - 5 = 6. From (2), we know that the hundreds digit is twice the tens digit. Say tens digit = x, thus, the hundreds digit = 2x.

=> x + 2x = 3x = 6 => x = 2 = An integer

Thus, the number is 425 and the product of the digits = 4*2*5 = 40. Unique answer. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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