Question 1190206: When a number is divided by 9, and the remainder is 6. What is the remainder when six times this number is divided by 9.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
x = some unknown integer
x/9 = quotient + (remainder/9)
x/9 = q + (r/9)
x = 9q + r
The items q and r will be integers as well.
The r can only take on integers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8}
Dividing this unknown number (x) over 9 leads to a remainder 6, so r = 6
x = 9q + r
x = 9q + 6
x = 3(3q + 2)
Now multiply both sides by 6
x = 3(3q + 2)
6x = 6*3(3q + 2)
6x = 18(3q + 2)
6x = 9*2(3q + 2)
Dividing both sides by 9 gets us
6x = 9*2(3q + 2)
6x/9 = 9*2(3q + 2)/9
6x/9 = 2(3q + 2)
Notice the 9's cancel on the right hand side.
We have some integer that represents the quotient with zero remainder.
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A numeric example:
The number x = 72 is a multiple of 9 because 8*9 = 72.
Add on 6 more to guarantee we have remainder 6
72+6 = 78
x/9 = 78/9 = 8 remainder 6
So 6x = 6*78 = 468
And 6x/9 = 468/9 = 52 remainder 0
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Answer: remainder = 0
Answer by ikleyn(52794)  (Show Source):
You can put this solution on YOUR website! .
A short version explanation
You are given
n = 9m + 6
(integer number n gives a remainder 6 when divided by 9).
Multiply both sides by 6. You will get
6n = 6*9*m + 36.
It can be re-written equivalently as
6n = = 9*(6m + 4),
which illustrates and proves that the number 6n is multiple of 9, i.e. is divisible by 9 with no remainder.
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