Bunuel wrote:
ngoctraiden1905 wrote:
I think it's easier: because all of the term divisible by 3 so the remainder of the multiply of them will be the remainder when after dividing by 3 then dividing by 11. This can only be 3 because all other choice is greater than 11
That's not right:
Consider this 6*6*6 divided by 33. What is remainder?
According to you logic: as all of them are divisible by 3 remainder must be less than 11, but in this case remainder is 18>11.
For the original question the answer IS 30. But my point was that the solution I provided is not easy, so I wonder if there is some easier way to do the same.
There is a *slightly* easier solution. If we factor out a 3 from each number, we come up with:
\(3*3*3*3*(348*349*310*311)/33\) = \(3*3*3*(348*349*310*311)/11\)
Now our task is to reduce each number to its remainder when divided by 11.
We notice that 352 is the closest multiple of 11 to the numbers inside the parentheses, and thus the remainders are (-1 * -2 * -3 * -4)* 3 * 3 * 3.
-4*-3 = 12; remainder 1
-1*2 = 2, remainder 2
3*3*3 = 27; remainder 5
Now 5*2*1 = 10, which is the remainder when divided by 11. Multiplying this by 3 (because the question asks the remainder when divided by 33), we get 10*3 = 30.
Answer: C
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