imhimanshu wrote:Can you please explain how to approach the below question -
A no. N when divided by divisor D leaves remainder 23. when same no. N is divided by 12D remainder is 104. what will be the remainder when the no is divided by 6D
Thanks
Given N/D = 23 => N = xD+23 -------------- equn 1
also, N/12D = 104 => N= 12yD+104 ------------equn 2
It is evident that N>=104
and D> 23, because the least remainder is 23.
From 1 & 2 we get
D (x-12y) = 81 = 27*3 = 81*1
Since D>23 we can two values for D i.e. either 27 or 81
Let's start with 27.
Since N>= 104 and 104/27 gives us 23 as the remainder, Lets Keep N= 104.
Let's use this value in equn 2 to check whether this satify the given condition.
12D= 27*12 = 324
104/324 gives 104 as the remainder.
Now 6D = 162
and 104/162 also gives 104 as the remainder.
Hence the remainder will be same in both the cases.
You can also do this-- add 104 to 324, this will give the value of N = 428.
Since N/D gives the remainder as 23. and we have the value of D=27, lets check it by using these values: N-23 should get divided by 23 i.e. 405
and yeah, 405/27 is divisible by 27. 27*15 = 405
I don't know whether this is the correct way to solve this question... I just saw the solution by Anurag and it makes much more sense.
