Solution

Well, first let me rephrase the question. Simply put, what I have stated is that there is a number, which when divided by 3 leaves a remainder of 2. The number when divided by 4 leaves a remainder of 1. When the number is divided by 7 leaves 4 as the remainder. What will be the remainder if the number is divided by 84 (the LCM of 3, 4 and 7)?

I have arrived at the solution in a very "un-mathematical" way. I am sure you have a better method.

This is what I did:

Let the number be "x". When it is divided by 3, we have 2 as the remainder. Let q1 be the quotient in this case. Therefore, x = 3q1 + 2.

When "x" is divided by 4, it leaves 1 as the remainder. In this case, let q2 be the quotient. Then, x = 4q2 + 1.

When "x" is divided by 7, a remainder of 4 is left. Let q3 be the quotient in this case. So, we have: x = 7q3 + 4.

Now, x = 3q1 + 2. For q1 = 0, 1, 2, 3, 4, 5, 6, ............, we see that x can be any of these:

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, ...........

x = 4q2 + 1,

So, x can be any of the following: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, ...........

Finally, x = 7q3 + 4,

So, x can be any of the following: 4, 11, 18, 25, 32, 39, 46, 53, 60, 67, ...........

The number has to satisfy all the three conditions; therefore, it is a number that must occur in all the three series.

A closer look at the three series shows that 53 is the first number that is common to all of them. So, the number could be 53. Now, the LCM of 3, 4, and 7 is 84. This means that 84 is the smallest number that is a multiple of all the three numbers: 3, 4, and 7. This implies that if 53 is common to all the three series, then 53 + 84 = 137 will also be common to all the three series. Not just that, the sum of 53 and all multiples of 84 will also be common to all the three series. Therefore, 53, 53 + 84, 53 + 2(84), 53 + 3(84), 53 + 4(84), etc., will all be common to the three series. So, the number x can be any of the following: 53, 137, 221, 305, ......

We have to find only the remainder when the number is divided by 84. That can be done by choosing any number and dividing by 84. The remainder will be the same in every case; the quotient will differ. Let's pick 305. When divided by 84, it leaves a remainder of 53. In fact, I could have chosen even 53, which is smaller than 84. When 53 is divided by 84, the quotient will be zero and the remainder will be 53. So, the remainder when the number is divided by 84 will be 53.

I am not really happy with this method. I was plain lucky to spot that the three series meet at 53. Any elegant solution that relies more on mathematics than on keen eyesight?