Question

Is there an integer that has a remainder of 2 when it is divided by 5 and a remainder of 3 when it is divided by 6? a. Is there an integer n such that n has ______? b. Does there exist _____ such that if n is divided by 5 the remainder is 2 and if ______?

Answer 1

Answer:

12314122

Step-by-step explanation:

Answer 2

Yes, that type of number exist and they are 27,57,87,117,147,177..and so on.

Step 1: Find the number.

Given -  Find an integer that has a remainder of 2 when it is divided by 5 and a remainder of 3 when it is divided by 6.

Let n be the number which satisfy the condition.

so, $n= 5p+2=6q+3$

When we divide  $6q+3$ by  5, we have remainder of  $q+3$, since  $6q+3=5q+(q+3)$  and  $5q$ is divisible by  5 .

Therefore, $q+3 = 2 (mod 5)$, and so,$q = -1=4 (mod 5) .$

With this, we can write q as a number of the form $5r+4$ .

Putting this into $n=6q+3$ , we have  $n=6$×$(5r+4)+3=30r+27$

If we divide  $30r+27$ by  5 , we get a remainder of  2: note that  $30r+27=30r+25+2=5(6r+5)+2$ . And if we divide it by  6 , we get a remainder of  3 by the same token,  $30r+27=30r+24+3=6(5r+4)+3$

So, our number will be: $27,57,87,117,147,177,207....$ and so on