Dear moderators and friends,

When positive integer n is divided by 6, the remainder is 2; when n is divided by 8, the remainder is 4.
What is the remainder when n is divided by 48?

a. 8
b. 16
c. 20
d. 26
e. 32

n = 6k + 2
n = 8k + 4

n = 48k + ?

I did the usual LCM, and it came to:
48K + 40.

So i went on to do the usual brute calculation:
n = 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68
n = 12,20,28, 36,44,52,60,68

1st common term > 48 is 68.
Thus remainder is 20.

Is there any way to solve it faster?
Can we use the equation n = 48K + 40??

Thanks
KH

Why do you want to get a number greater than 48.
Even the smallest such number number 20 is fulfilling the criteria.

Divided by 6 it leaves 2, divided by 8 it leaves 4, divided by 48 it leaves 20.

Makes sense at saves time too.

Remember that question says number has to be +ve. But it has to be greater than 48 or not is not given/asked. So there is no need to make calculation lengthy on our own.

KH,
I find it easier if you just try the answer. And see if they fit the condition of divided by 6 remain 2 and divided by 8 remain 4.

Say 48+32=80 , it doesn't fit . It is hard to guess it with 20 I think ?

Maria

please cite the author of this question or we will have to delete it!