Using ingenuity on remainder questions

We know various types of  remainder questions and discussed how to tackle them. You may also have examined the concepts of general divisibility and divisibility specifically applied to remainder. There is one concept, however, that we haven’t discussed yet, and that is using ingenuity on remainder questions. Say “x” gives you a remainder of 2 when divided by 6. What will be the remainder when x + 1 is divided by 6? Go back to the divisibility concepts discussed above. When x balls are split into groups of 6, we will have 2 balls leftover. If we are given 1 more ball, it will join the 2 balls and now we will have 3 balls leftover. The remainder will be 3. What happens in the case of x + 6 – what will be the remainder when this is divided by 6? This additional 6 balls will just make an extra group of 6, so we will still have 2 balls leftover. What about the case of x + 9? Now, of the extra 9 balls, we will make one group of 6 and will have 3 balls leftover. These 3 balls will join the 2 balls leftover from x, giving us a remainder of 5. Now, what about the case of 2x? Recall that 2x = x + x. The number of groups will double and so will the remainder, so 2x will give us a remainder of 2*2 = 4. On the other hand, if x gives us a remainder of 4 when divided by 6, then 2x divided by 6 will have a remainder of 2*4 = 8, which gives us a remainder of 2 (since another group of 6 will be formed from the 8 balls). Let’s consider the tricky case of x^2 now. If x gives us a remainder of 2 when it is divided by 6, it means: x = 6Q + 2 x^2 = (6Q + 2)*(6Q + 2) = 36Q^2 + 24Q + 4 Note here that the first and the second terms are divisible by 6. The remainder when you divide this by 6 will be 4. We hope you understand how to deal with these various cases of remainders. Let’s take a look at a  sample question now: If z is a positive integer and r is the remainder when z^2 + 2z + 4 is divided by 8, what is the value of r? Statement 1: When (z−3)^2 is divided by 8, the remainder is 4. Statement 2: When 2z is divided by 8, the remainder is 2. This is not our typical, “When z is divided by 8, r is the remainder” type of question. Instead, we are given a quadratic equation in the form of z that, when divided by 8, gives us a remainder of r. We need to find r. This question might feel complicated, but look at the statements – at least one of them gives us data on a quadratic! Looks promising! Statement 1: When (z−3)^2 is divided by 8, the remainder is 4 (z – 3)^2 = z^2 – 6z + 9 We know that when z^2 – 6z + 9 is divided by 8, the remainder is 4. So no matter what z is, z^2 – 6z + 9 + 8z, when divided by 8, will only give us a remainder of 4 (8z is a multiple of 8, so will give remainder 0). z^2 – 6z + 9 + 8z = z^2 + 2z + 9 z^2 + 2z + 9 when divided by 8, gives remainder 4. This means z^2 + 2z + 5 is divisible by 8 and would give remainder 0, further implying that z^2 + 2z + 4 would be 1 less than a multiple of 8, and hence, would give us a remainder of 7 when divided by 8. This statement alone is sufficient. Let’s look at the second statement: Statement 2: When 2z is divided by 8, the remainder is 2 2z = 8a + 2 z = 4a + 1 z^2 = (4a + 1)^2 = 16a^2 + 8a + 1 When z^2 is divided by 8, the remainder is 1. When 2z is divided by 8, the remainder is 2. So when z^2 + 2z is divided by 8 the remainder will be 1+2 = 3. When z^2 + 2z + 4 is divided by 8, remainder will be 3 + 4 = 7. This statement alone is also sufficient. Because both statements alone are sufficient, our answer is D.