Mixed Problems: Set 3 (Greatest Terms, Remainders, Multinomials)

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge to start. Take one plus x plus x squared, to the power six. I want the coefficient of x to the power four. Will you multiply out all six brackets? Each bracket has three terms, so that is three to the power six: seven hundred and twenty nine products! And in the exam you have about two minutes, right? Today is Set three of our mixed problems: greatest terms, remainders and multinomials. Each type has one tool, and the whole game is to spot it fast. These questions come up often in JEE Main, so this speed really matters. Let's go.

0:43Tool 1: remainders, just next to a multiple

So, first tool: remainders. We want the remainder when a to the power N is divided by m. The trick: write a, or a small power of a, as k m plus or minus one, where k is an integer. The sign with plus above minus is read, plus or minus. Okay? Now think. Expand k m plus one, to the power N, with the binomial theorem. Which terms carry a factor of m? All of them except the last one, yes, exactly! With N a positive integer, the last term is one to the power N, which is just one. So k m plus one, to the power N, equals m times an integer, plus one. The remainder is one. Clear? And for k m minus one, only the last term changes. It becomes minus one to the power N, nothing else. So we get m times an integer, plus the last term, minus one to the power N, got it? Now, suppose N is odd. Then the last term is minus one. So the remainder is minus one, right? Oh no, careful! Remainders are never negative. Remainder minus c means the true remainder is m minus c: just add m once. So minus one becomes m minus one. With me?

2:02Tool 2: the greatest term is not always the middle one

Alright, second tool: the greatest term. First, recall: when n is even, the greatest binomial coefficient is the middle one, n C, n over two. On screen, the small n sits at the top left of the capital C, and n over two sits at its bottom right. Okay? So the middle term is always the greatest term, right? No, no, wait! The binomial coefficient is only one part of a term. The term also carries powers of x and of a. So for x plus a, to the power n, with given numbers x and a, and n a positive integer, find a number called m. Numerically greatest means greatest in size, ignoring the sign. m equals n plus one, times mod a, over, mod x plus mod a. The vertical bars mean the modulus, the size without the sign. Clear? If m is not an integer, the numerically greatest term is T of, the integer part of m, plus one. Capital T with a small label below it is that term of the expansion, and the square brackets mean the integer part. Never forget that plus one! Got it? And if m is exactly an integer, then T m and T m plus one are equal in size, and both are the greatest. Write both of them, or the answer is incomplete. Remember that!

3:24Tool 3: multinomials, one pick from each bracket

Now, third tool: multinomials, brackets with more than two terms. x one, x two, up to x k are k independent letters; the small numbers below the x are just labels, and the three dots mean, and so on. With n a positive integer, the number of terms is n plus k minus one, C, k minus one. Okay? But often, the terms are powers of one letter, x. Say, the coefficient of x to the power m in a plus b x plus c x squared, to the power n. How do you build x to the power m from these n brackets? One pick from each bracket, yes! Pick a in p brackets, b x in q brackets, and c x squared in s brackets. There are n brackets, so p plus q plus s equals n. And the power of x is q plus two s, which must equal m. Right? Each such triple p, q, s, in round brackets, gives n factorial, over p factorial, q factorial, s factorial; the exclamation mark means factorial. Times a to the power p, b to the power q, and c to the power s. Then add over every triple. And careful: different triples can give the same power of x, so never stop at the first one! With me?

4:49Example 1: remainder when 2^60 is divided by 7

Now, Example one, an easy one from Set three. Find the remainder when two to the power sixty is divided by seven. Start your clock, pause the video, and try it. Done? Okay, let's check. Two itself is not next to a multiple of seven. So which small power of two is? Two cubed! Yes, exactly. Two cubed is eight, and eight is seven plus one. Perfect! Now, sixty is three times twenty. So two to the power sixty is two cubed, in brackets, to the power twenty. That is eight to the power twenty, which is seven plus one, to the power twenty. Okay? Now expand it. Every term carries a factor of seven, except the last one, which is one. So it is seven times an integer, plus one. The remainder is one! Under a minute, and done. Clear?

5:46Example 2: coefficient of x^4 in (1 + x + x^2)^6

Now, remember our challenge from the start? Let's finish it. Example two, a medium one, also from Set three. Find the coefficient of x to the power four in one plus x plus x squared, to the power six. Start your clock, pause the video, and try it. Done? Okay, let's check. Third tool. Pick one in p brackets, x in q brackets, and x squared in s brackets. Six brackets, so p plus q plus s equals six. And the power of x, q plus two s, must be four. Right? Start with s equal to zero, no x squared at all. What are q and p then? q is four, and p is two. Correct! So the triple is two, four, zero. Six factorial over two factorial, four factorial, zero factorial; zero factorial is one. Seven hundred and twenty over forty eight: fifteen. Okay? So the coefficient of x to the power four is fifteen, right? No, no, careful! That is the trap. x squared also builds powers of x. Take s equal to one: then q is two, and p is three. Six factorial over three factorial, two factorial, one factorial: seven hundred and twenty over twelve, that is sixty. Got it? And s equal to two: q is zero, and p is four. Six factorial over four factorial, zero factorial, two factorial: seven hundred and twenty over forty eight, fifteen again. Clear? Now add them up: fifteen plus sixty plus fifteen. Ninety! See, no seven hundred and twenty nine products. Just three triples, and done. Perfect!

7:44Quick revision: remainders, greatest term, multinomials

So, quick revision. Remainders: write a small power of the base as k m plus or minus one, and never leave a negative remainder. Greatest term: find m; the answer is T of the integer part of m, plus one, or T m and T m plus one tie when m is an integer. Multinomials: list every triple p, q, s with p plus q plus s equal to n, and q plus two s equal to m, then add. Now attempt the whole of Set three with your clock running, and then try the practice problems on this page. You are ready for this. All the best!