SOLUTION: A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C

Question 1209434: A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52908)   (Show Source): You can put this solution on YOUR website!
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A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C
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Since 4 * AB = CA, CA must be a two-digit number >= 40. (*) Since 4 * AB = CA, we conclude that AB is a two-digit number between 10 and 24, inclusive. So, A is either 1 or 2. But from 4 * AB = CA, digit A must be even - so, we conclude that A = 2. Thus, CA is a two digit number multiple of 4 and with the last digit 2. Such numbers CA can be only 12, 32, 52, 72, 92. Condition (*) prohibits 12 and 32, so the remaining opportunities for CA are 52, 72, 92. Divide CA by 4, and you will get the values 13, 18, 23 for AB. But, as we established above, the digit A is 2; therefore, AB can only be 23. Thus, A = 2; B = 3; C = 9. All these digits are different. The sum A + B + C is 2 + 3 + 9 = 14. ANSWER

Solved.

Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!

A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C 4AB = CA Obviously, AB and CA are 2-digit numbers As such. AB = 10A + B, and CA = 10C + A. So, 4AB = CA becomes: 4(10A + B) = 10C + A 40A + 4B = 10C + A 4B = 10C + A - 40A 4B = 10C - 39A B = Now, we can see that C - 39A is a MULTIPLE of 4, so A CANNOT be an ODD DIGIT. Furthermore, the digit A MUST be less than 3, as ANY digit greater than 3 would yield a negative (< 0) numerator. And, neither can digit A be 0 because, while the units digit in CA can be 0, the tens digit in AB CANNOT. Having said that, digit A being less than 3, not ODD, and NOT 0 makes it 2 B = B = ----- Substituting 2 for A B = At this point, 10C MUST be greater than 78, which means that C can be 8 or 9. But, if C = 8, then numerator becomes 2, and 2 is NOT a MULTIPLE of 4. So, C MUST be 9 B = ------ Substituting 9 for C B = B = B = 3 Therefore, (A, B, C) = (2, 3, 9), and A + B + C = 2 + 3 + 9 = 14 You can do the CHECK!!

Answer by greenestamps(13215)   (Show Source): You can put this solution on YOUR website!

A, B, and C are distinct digits and 4* AB = CA.

The given multiplication puts two restrictions on the choice for digit A:
(1) 4 * AB gives a result with units digit A. 4 times any integer yields an even product, so A must be even.
(2) 4 * AB gives a result that is a 2-digit integer. That means A must be no greater than 2.

Those two conditions together mean A must be 2.

Now 4 * AB yields a product with units digit 2, so be can only be either 3 or 8.

But 4 * 28 yields a product of 3 digits, so B must be 3.

So we have that the indicated multiplication must be 4 * 23 = 92.

So A = 2, B = 3, and C = 9.

ANSWER: A+B+C = 2+3+9 = 14

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