When dividing the number "a" by 9, the remainder is 5. What condition must the number "b" satisfy in order for the difference a-b to be a multiple of 9?

Nadia Smith 2022-09-13 Answered
When dividing the number "a" by 9, the remainder is 5. What condition must the number "b" satisfy in order for the difference a-b to be a multiple of 9?
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Answers (1)

tranciarebt
Answered 2022-09-14 Author has 10 answers
When dividing the number "a" by 9, the remainder is 5
So the number a can be written as
a=9k+5
Now let's use the divisibility property:
"If both the minuend and the subtrahend are divisible by some number, then the difference is also divisible by that number"
write down our difference
a-b=9m(9k+5)-b=9m
note that the number b can also be divided by 9 with a remainder
so let's write it as
b=9n+x
and now our difference will look like this
a-b=9m(9k+5)-(9n+x)=9m9(k-n)+(5-x)=9m
so that this equality holds x=5
And then the number b must be divisible by 9 with a remainder of 5
let's give an example:
50:9= 5*9+5
41:9=4*9+5
50-41=9 and it's a multiple of 9
***************
221:9=24*9+5
140:9=15*5+5
221-140=81
and it is a multiple of 9

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