MATHCOUNTS NATIONAL 2014 #12

Question

I don't know how to approach this. The correct answer is $2013$.

What is the value of

$$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}\,?$$

I thought about using Binomial Theorem in some way as the format looks similar to the results you get, but couldn't follow through with it.

here is a mathjax tutorial written for beginners. Please go through the same and try to typset your question using mathjax instead of links to external source. Also it would be great if you can provide your own ideas for solving this problem. — Paramanand Singh, Mar 16 '21 at 02:14
You can try to identify the expression with typical formula from school algebra. Can you do that? — Paramanand Singh, Mar 16 '21 at 02:17
Have you studied formulas for $(a+b) ^2,(a-b)^2,(a+b)^3,(a-b)^3$ in your school? Try to replace 2013 by $a$ and $2014$ by $b$ and see if you can find any formula similar to the expression in question. — Paramanand Singh, Mar 16 '21 at 02:21
You should also note that we don't need binomial theorem here. The binomial theorem extends such formula for larger powers. — Paramanand Singh, Mar 16 '21 at 02:24
I see no legitimate grounds for closing the question: the OP did in fact provide context (the tag) and evidence of having thought about the question and made a relevant observation. — Brian M. Scott, Mar 16 '21 at 05:54

score 3 · Accepted Answer · answered Mar 16 '21 at 02:19

3

Your idea of using the binomial theorem is a good one, but there’s a little more work to be done. Most of the numerator looks a lot like the result of expanding $(2013-2014)^3$, so you should start by writing out that expansion. Then you can easily see that the numerator is

$$(2013-2014)^3+2013^2\cdot 2014+1\,;$$

Can you finish it from there?

answered Mar 16 '21 at 02:19

Brian M. Scott

616,228

Wow! Yes, I can. May I ask what you saw in order to realize that was the case? – Smartsav10 Mar 16 '21 at 02:20
@Smartsav10: I just looked at the expansion of $(2013-2014)^3$ and asked myself what adjustment needed to be made to get the numerator; and when I wrote that down, it all fell into place. – Brian M. Scott Mar 16 '21 at 02:21
Thank you! I'll remember to start including expansions of things in my thought process. – Smartsav10 Mar 16 '21 at 02:25
1

@Smartsav10: You’re welcome! – Brian M. Scott Mar 16 '21 at 02:28
2

I’d really like to know the justification for the downvote. – Brian M. Scott Mar 16 '21 at 05:52

score 3 · Answer 2 · answered Mar 16 '21 at 02:21

3

Consider the expression $$\frac{a^3 - 2a^2b + 3ab^2 - b^3 + 1}{ab}$$ where $a = 2013$ and $b = a+1 = 2014$

Put $b = a+1$ and simplify!

$$\frac{a^3 - 2a^2b + 3ab^2 - b^3 + 1}{ab} = \frac{a^2(a+1)}{a(a+1)} = a = 2013$$

answered Mar 16 '21 at 02:21

PTDS

3,464

While this works, I think for contest math it will be better to go with @Brian M. Scott’s solution – Smartsav10 Mar 16 '21 at 02:29

MATHCOUNTS NATIONAL 2014 #12

2 Answers2