If a positive integer m is divided by d, the remainder is 5. What is

How da can be same for both the expressions.

m= da+5
m/3 = da+15

Could you explain this?

gvij2017

Well i am not completely sure.of the solution.
I solved question considering a to be same..




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IMHO, the question implicitly asks to solve a system of (max) 3 equations in 4 variables. Impossibile by definition.

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Albymana , Yes It's impossible to solve the Question.

Dear Bunuel , Please post OA As soon as possible.

Waiting for It!!!!!!!!!!!

Unsure of my answer but here goes my reasoning.

Question statement analysis gives

m = kd + 5 where k is an integer greater than or equal to 0.
Since divisor is alway greater than the remainder, d > 5.


Statement 1 analysis: (m/3)/d = qd + 15

Dividing "m/3" by d is same as multiplying m/3 by 1/d.
Thus the statement tells us that "m" divided by "3d" gives a remainder of 15.
Again, since divisor is greater than the remainder, the above statement implies 3d>15 which means d>5.

We already knew this from the question statement analysis. Thus, statement 1 is insufficient.



Statement 2 analysis:

D>20

No info on m or d hence insufficient.


Combining statement 1 & 2 gives us no new information about either m or the range of d except that d>20. Thus we can deduce no concrete value of d.
Insufficient.

The answer is E

You are wrong in the highlighted portion..
say m=45 and d=20..
45 divided by 20 gives 5 as remainder
Now 45/3 or 15 divided by 20 will give remainder as 15.
BUT m divided by 3d, that is 45 divided by 3*20 or 60 will give 45 as remainder.
so m/3 divided by d and m divided by 3d are different when you are checking the remainders

m/d = R(5 ) means d>5 and m = d.x + 5 , where x is any integer

1.
let m/3 be y
---> y = d.p + 15 where p is any integer . Also , note that d>15 otherwise we cant generate remainder of 15.
substitute from stem the value of m

---> dx + 5 = 3dp + 45
d = 40 / ( x- 3p )

Now lets find some values of d so that we can prove this statement insufficient.
x = 4 p = 1 so, d = 40
x = 5 p= 1 so, d = 20
Hence , insufficient
B--> insufficient

Combined

d = 40 only
Sufficient

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