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Positive integer x leaves a remainder of 2 when divided by 8. What is the remainder when x+9 is divided by 8

 Aug 21, 2016
 #1
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There is not one UNIQUE solution but an infinite number of them. Here is the general form of the solution: x =2 + 8c, where c is any positive integer. So, the numbers would start with:

2, 10, 18, 26, 34......etc.

 

The 2nd part of your question depends on what value you take.

So, [2 + 9] mod 8=3. And so the remainder will be 3, no matter what value you choose.

 Aug 21, 2016
 #2
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Previous guest is wrong. Answer is 25/8

 

Solution

 

Let x /8 = Q  + 2    so  x = 8(Q +2)   where Q is the quotient and 2 is the remainder

 

then   x  + 9  = 8(Q +2)  +9   so

(x  +9) /8 = {  8(Q +2) +9 }  /8

              = Q +2 +9/8

 

which is the quotient plus a remainder of 25/8 or 3 plus a quarter.

 Aug 21, 2016
 #3
avatar+37160 
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Guest #2 is wrong:

 

"x /8 = Q  + 2    so  x = 8(Q +2)   where Q is the quotient and 2 is the remainder"       is INCORRECT

                                x = 8(Q)  + 2   is correct

 

 

anyway , if you start with a reaminder of  2   and add 9  its like a remainder of 11....pull another '8' out of there for a remainder of '3'

 Aug 21, 2016
 #4
avatar+9674 
+5

x(mod8)2x+8(mod8)2x+9(mod8)3

Guest #2 is actually wrong. Note that x is an positive integer and 8 is also a positive integer. The remainder won't be a fraction.

 

This could be solved using modulus maths like Guest #1 and I did.

 Aug 22, 2016
 #5
avatar+9674 
0

Therefore x+9/8 = something with remainder 3.

MaxWong  Aug 22, 2016
 #6
avatar+26397 
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Positive integer x leaves a remainder of 2 when divided by 8.
What is the remainder when x+9 is divided by 8

 

x2(mod8)x+92+9(mod8)x+911(mod8)x+9118(mod8)x+93(mod8)

 

laugh

 Aug 23, 2016

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