# Putnam for Amusement: 1998 B1 and 1991 A2

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• Published on Oct 2, 2018
• Let's have some fun with slick algebraic manipulations.
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• How it feels like 5 months ago

When you get 3(x+1/x) in the first problem you can make a quadratic function by letting the expression equal to a constant...then to get real values you need the delta x (b^2-4ac) >= o so then you get that the expression will be greater than or equal 6 ๐

• puzzle collector 6 months ago

wow this is fun!

• Mohd Tauheed Khan 7 months ago

Put x=1

• Mohd Tauheed Khan 7 months ago

I used minimum value of (x+1/x) =2

• Amber Deshbhratar 11 months ago +3

this is a very good USclip channrl

• Arvind Kumar 11 months ago

The first one has also come in Regional Mathematics Olympiad of India also

• Paco Libre 11 months ago

If you know paschalโs triangle, expanding the 1998 problem isnt so bad. [x^6+6x^4+15x^2+20+15x^-2+6x^-4+x^-6] and [x^3+3x+3x^-1+x^-3] I have the fourth row memorized, and getting to the sixth row is just some addition.

• etotheipi 11 months ago

I'm actually surprised that these are Putnam problems. I've always viewed Putnam problems as USAMO/IMO level, but I would have thought as the first one as an AMC 10 difficulty and the second one as easier than a JMO 1. I guess the Putnam has gotten a lot harder over the years.
Thanks a lot for making these vids! They're well explained as always.

• Felipe Lorenzzon 11 months ago

I tried to add the functions inside each parenthesis on the first question, but it didn't help me out

• Who does Putnam problem for amusement lol.

• I did the first one in a different i think easier way.
I said let x0 be the value for which this expression is minimal.
Then i said lets substitute t=1/x0, which we can because x0!=0.
It's easy to see that the expression remains the same for t=1/x0 and has therefore the same value x0 for to substitute instead of t to make it minimal so for t=x0 the expression is minimal but t=1/x0 so x0=1 for the minimal value of the expression, substituting in the expression gives you 6.

• Luka Mernik 10 months ago +4

I don't think this is correct. You just got lucky. In general a function does not have a unique minimum, so x_0 and 1/x_0 can be two different minimums and are not necessarily the same point. I.e. if the function has a minimum at 2 and 1/2, then the value f(2)=f(1/2), but x_0 and 1/x_0 are not equal.

• Randall Kim 11 months ago

I have a request for a problem. Please take a look at the 2006 AIME II (Alternate AIME) #13. Keep up the great videos!

• adandap 11 months ago +2

I like the first problem. I wasn't quite as slick as you, but the binomial approach isn't a dead end. You can use is recursively to express x^6 + x^(-6) in terms of 4th and 2nd powers, and then keep going until everything is in terms of powers of (x+1/x). You end up in the same place. (I hope that was clear.) Now onto the other one...

• Ryan C. 11 months ago +1

This is weird, the first problem was on the 1998 Putnam (mks.mff.cuni.cz/kalva/putnam/putn98.html), but the exact same problem was also on the 1979 AHSME (artofproblemsolving.com/wiki/index.php?title=1979_AHSME_Problems/Problem_29). I find it really strange that the same exact problem would be on tests of incredible different difficulty and also tests nearly 20 years apart.

• el tapa 11 months ago +1

Oh mate, amazing

• 77Chester77 11 months ago +30

As always: At the beginning I have no clue at all but at the end you made it look so easy. Bravo.

• Nikolajus Elmutis 11 months ago

How comes they are so easy u.u

• Sarthak Sharma 11 months ago

Sir 1998 question
Can be simply solved by using AM>=GM inequality

• Avishek Tiwari 11 months ago

• Jeff Ahn 11 months ago

Woow. So elegant.

• Siddhartha Shree kaushik 11 months ago +2

I got 3 notifications within 2 minutes
Are you serious?

• Siddhartha Shree kaushik 11 months ago