Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take five to the power two thousand and twenty six. That number has one thousand four hundred and seventeen digits. Your calculator cannot even display it. Now divide it by thirteen. What is the remainder? Impossible, right? By the end of this class, you will find it in about thirty seconds, without a calculator. This is Part A of our binomial theorem problem bank, at JEE Main level, and remainders like this come up often. So, let's begin.

0:35Part A: a 40-minute test

So, here is what Part A holds. Twenty solved problems, all at JEE Main level. Short, sharp questions, okay? And they are mixed: terms, sums of coefficients, remainders, greatest terms and multinomials, all together. Nobody tells you which chapter a question comes from, just like the real exam. Right? The plan: finish Chapters one to seven, then attempt Part A as one forty minute test. So today, two tools that some Part A problems need, then two of its problems, one easy and one harder. With me?

1:11Tool 1: remainders from the binomial theorem

Tool number one, from Chapter five: remainders. We want the remainder when a to the power capital N is divided by m. The capital N, written small at the top right, is the power, and it must be a positive integer. Okay? The trick: write a, or a small power of a, as k times m, plus or minus one. The sign on screen, plus above minus, means plus or minus. And k is just an integer. Right? Now think. Suppose we expand k m plus one, the whole to the power N, by the binomial theorem. Which terms of that expansion contain m? Every term, except the last one! Yes, exactly! And the last term is one to the power N, which is just one. So, k m plus one, in brackets, to the power N, equals m times an integer, plus one. Remainder one. Clear? Now put minus one inside. The last term becomes minus one, to the power N, so k m minus one, in brackets, to the power N, equals m times an integer, plus, in brackets, minus one, to the power N, as shown. When N is even, that is plus one; when odd, minus one. Got it? But wait. When N is odd, the remainder seems to come out as minus one. Can a remainder really be negative? No, it cannot! Remainders are never negative. So a remainder of minus c really means m minus c: just add one m, and you have the true remainder. Remember this line; we will need it very soon. With me?

2:46Tool 2: sums of coefficients

Tool number two, from Chapter four: sums of coefficients. Take any polynomial, f of x. The idea is really simple: put convenient values of x. Okay? Now tell me. Which value of x makes every power of x equal to one? Yes, exactly: put x equals one! Then every term becomes just its coefficient. So the sum of all the coefficients is f of one. Right? And now put x equals minus one. What happens to the odd powers of x? They turn negative. Correct! Even powers stay plus, odd powers turn minus. So add f of one and f of minus one: the odd ones cancel, and the even ones come twice. Divide by two, the fraction bar on screen, and you get the sum of the coefficients of the even powers. Clear? And subtract instead: f of one, minus f of minus one, over two. Now the even ones cancel, and you get the sum for the odd powers. Two values of x, and three sums. That's it! Got it?

3:54Example 1: remainder when 5^2026 is divided by 13

Now, remember our puzzle? Let's finish it. Find the remainder when five to the power two thousand and twenty six is divided by thirteen. Pause the video, and try it first. Hint: try a small power of five. Done? Okay, let's check. Five squared is twenty five, and twenty five is twenty six minus one. And twenty six is two times thirteen; the cross means times. So five squared is in our form, k m minus one, with m equal to thirteen. Perfect! Two thousand and twenty six is twice one thousand and thirteen. So five to the power two thousand and twenty six is five squared, in brackets, to the power one thousand and thirteen. That is twenty six minus one, to the power one thousand and thirteen. With me? Now use tool one. Every term of this expansion has twenty six in it, and so thirteen too, except the last term: minus one, to the power one thousand and thirteen. So, is that last term plus one, or minus one? Minus one, yes! One thousand and thirteen is odd. So we get thirteen k, plus, in brackets, minus one, to the power one thousand and thirteen; that is, thirteen k minus one, where k is an integer. Right? So, the remainder is minus one, right? No, no, careful! Remainders are never negative. This is a very common trap. Take one thirteen out of thirteen k, and give it to the minus one. Thirteen k minus one equals thirteen times k minus one, in brackets, plus twelve. Clear? So the remainder is thirteen minus one: twelve. Exactly what tool one said: minus c means m minus c. One thousand four hundred and seventeen digits, and we never needed even one of them! See how fast that was?

5:44Example 2: a2 + a4 + ... + a12 for (1 + x - 2x^2)^6

Now a slightly harder one. One plus x minus two x squared, the whole to the power six, is expanded as one, plus a one x, plus a two x squared, and so on, up to a twelve x to the power twelve. Here a one, a two and the rest, with small numbers at their feet, are just the coefficients, and the dots mean and so on. Find a two plus a four, and so on, up to a twelve. Pause the video, and try it. Hint: use tool two. Done? Okay, let's check. Call the whole left side f of x. Put x equals one: one plus one minus two is zero, and zero to the power six is zero. So f of one is zero. Okay? Now put x equals minus one: one minus one minus two is minus two. Minus two, to the power six: sixty four. An even power, so it comes out positive. Right? So, by tool two, the coefficients of the even powers add up to f of one, plus f of minus one, over two. Zero plus sixty four, over two: thirty two. Got it? So the answer is thirty two, right? Hmm, wait. Tool two adds every even power. Does the sum in the question have every even power? No, no, wait! The constant term, one, is x to the power zero, and zero is even. So thirty two is one, plus a two, plus a four, and so on, up to a twelve. But the question never asked for that one. Clear? So take it away: a two plus a four, and so on, up to a twelve, equals thirty two minus one: thirty one. Many students stop at thirty two. Always check whether the constant term belongs to your sum. Perfect!

7:31Quick revision: remainders, f(1) and f(-1), constant term

So, quick revision. One: for remainders, make the base k m plus or minus one, expand, and turn a remainder of minus c into m minus c. Two: the sum of all the coefficients is f of one, and the even powers give f of one, plus f of minus one, over two. Three: read the asked sum carefully; is the constant term in it? Now go and attempt all twenty problems on this page as one forty minute test. You will do well, I promise. All the best!