Numbers and

Question

a number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor the remainder is 11. What is the value of the divisor? 

 

Answer 1

This is a great question. 

let's start out and say X is the dividend and d is the divisor For the 1st part. Then let's assume when we take X/d=k+ the remainder. (K is not important we are just using it as a place holder for now). The remainder is 24 And the remainder is actually a fraction so it can be written as 24/d. 

So again

x/d=k+24/d

 

now the second part. X is now 2x, d stays the same and k becomes another number (again not too important call it l) plus the remainder of 11. 

 

2x/d=l+11/d

 

now multiply both equations by d to get rid of fractions. 

 

We have the  equations as 

x=kd+24 

2x=ld+11

now multiply the 1st equation by 2 and set them equal to each other. 

 

2kd+48=ld+11

 

bring the constants to 1 side and the d to the other and factor out a d

 

37=d(l-2k)

heres the part where it gets a bit tricky. 

 

2 numbers multiply together to get 37 and one of the numbers is d. 

 

37 is prime so that means that d is either 1 or 37 (bc the problem is designed using integers). Since it obviously can't be 1, bc then there wouldn't be a remainder, d must be 37. 

 

So so as a check, let's assume k is 1, then l must be 3 bc (l-2k must b 1). 

So if d is 37 and k is 1  X must be 37+24 or 61. 

So x is 61 then 2 X is 122 and 122/37 = 3 remainder 11 so it does check.