Question 1189473: A,B, and C are distinct digits, and 8•AB=ACA . Find the product of A•B•C
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
A,B, and C are distinct digits, and 8•AB=ACA . Find the product of A•B•C
~~~~~~~~~~~~~~~
My response is not full Math solution, it is only one possible answer, found by trial and error.
The numbers are AB = 29, ACA = 232; the digits are A = 2, B = 9, C = 3.
The product A*B*C = 2*9*3 = 54. ANSWER
The reasons are that ACA is the 3-digit palindrome multiple of 8, and the smallest such a palindrome is 232.
In any case, my numbers and my digits satisfy imposed conditions.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
8 times "AB" is equal to "ACA".
Since 8 is even, the last digit of the product, A, must be even.
So the possibilities are
8*(2B)=2C2
8*(4B)=4C4
8*(6B)=6C6
8*(8B)=8C8
The only one of those that has any possible solutions is the first. So A is 2. (8 times 4B is at most 8*49<400; and similarly for A=6 or A=8....)
So 8*(2B)=2C2; and the product 8*(2B) is at least 8*20=160 and at most 8*29=232.
Finally, 8*29=232 is the only multiple of 8 greater than 200 and less than or equal to 232 that has units digit 2. So 8*29=232 is the solution to the given multiplication problem.
ANSWER: A*B*C=2*9*3=54
|
|
|
