Question

Dividing a number by 3, 5, 6, respectively, leaves 2, 4 and 4, respectively, what is the remainder of dividing that number by 105?​

Answer 1

Let, that number be x.

So, (x - 2) is completely divisible by 3.

And, (x - 4) is completely divisible by 5 and 7 both.

Now, LCM (5, 7) = 35; and, numbers completely divisible by both 5 and 7 are:

35, 70, 105, 140, 175, 210, 245, 280,………

Let's experiment with the above numbers now:

(35 + 4) = 39…… it is completely divisible by 3.

(70 + 4) = 74……. it leaves a remainder of 2, if divided by 3.

(105 + 4) = 109…… it leaves a remainder of 1, if divided by 3.

(140 + 4) = 144.… it is completely divisible by 3.

(175 + 4) = 179…. it leaves a remainder of 2, if divided by 3.

(210 + 4) = 214…. it leaves a remainder of 1, if divided by 3.

(245 + 4) = 249…. it is completely divisible by 3.

(280 + 4) = 284…. it leaves a remainder of 2, if divided by 3.

So, from the experiment involving above numbers, we get three numbers which qualify; these are 74, 179 and 284. All 74, 179 and 284 leave same remainder of 74, if divided by 105.

If we lengthen our experiment we are sure to get many more numbers, but with one common feature that each leaves a remainder of 74, if divided by 3.

Therefore, the answer is 74.