if a positive integer x is divided by 32,the remainder is 25.which of the following could be the remainder when 2x is divided by 12?

Question

sorry sir I forget to write positive integer x

Answer 1

 

x=32n+25 for n=1,2,3,4,5...

2x=2(32n+25) for n=1,2,3,4,5...

2x=64n+50 for n=1,2,3...

(64n+50)/12=(64n/12)+(50/12)

50/12 will always leave a remainder of 2

64n/12 for n=1,2,3,4,5,6...

For how many multiples of 64 will the remainders start to repeat ?

64/12=5 R 4

(64+64)/12=5 R 4 +5 R 4 =10 R 8

(64+64+64)/12=5 R 4 +5 R 4 + 5 R 4=15 R 12=16 R 0

For each 64 there is a remainder of 4 and there are three 4's in 12.

Therefore after the third multiple of 64, when you divide them by 12, the remainders will repeat(4,8,0,4,8,0,4,8,0...).

The remainders are 4+2=6, 8+2=10, and 0+2=2 which was already shown.(the 2's come from 50/12=4 R2)

This is just another way to solve the problem.

 

 

Answer 2

 x/32 = C + 25/32

Where C is some constant integer 1, 2, 3, 4 etc.

x = 32C + 25

2x = 64C + 50

2x/12 = 64C/12 + 50/12

2x/12 = 5C + 4 + (4C/12 + 2/12)
2x/12 = 5C + 4 + (4C + 2)/12

For all values of C, it seems to rotate between 2, 6, and 10 for possible remainders.

 

 

Do you have a list of possible answers?



"Which of the following"?
 

Answer 3

Hmmm.  the format of proving the answer is going to depend on the class you're in.  I'm relying on common sense, supported by memory of mathematical "rings"... discrete math maybe?

 

The remainder of 2x divided by 32 is going to be (25+25-32) = 18.

 

The remainder of 2x divided by 12 is going to be even and in the range (0,2,4,6,8,10)

 

Can we eliminate any of them? 

 

Okay, let's play with x. 

 

16 is exactly half of 32. Anything evenly divisible by 32 is evenly divisible by 16.  The remainder of x divided by 16 is going to be 26-16 = 9.

 

8 is exactly half of 16.  Under the same logic, the remainder of x divided by 8 is going to be 9-8=1.

 

So, we know that the remainder of x divided by 8 is going to be 1. 

 

Therefore, the remainder of 2x divided by 8 is going to be 2.

 

Now, if we consider how 8's and 12s can line up, every two 12s is three 8's, so there are only 3 possible remainders.  If the 8 and the 12 line up, then we have a remainder of 2.  If there is one extra 8, the remainder is 10.  If there are two extra 8s, the remainder is 10+8-12 = 6.

 

So, the possible remainders are (2, 6, 10).   That's your answer.

 

 

Unfortunately, not knowing what your class is, I'm not sure how much of this will be of value to you.