# Putnam for Amusement: 1998 B1 and 1991 A2

Embed

- Published on Oct 2, 2018
- Let's have some fun with slick algebraic manipulations.

Your support is truly a huge encouragement.

Please take a second to subscribe in order to send us your valuable support and receive notifications for new videos!

Every subscriber and every like are wholeheartedly appreciated.

How it feels like5 months agoWhen 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 collector6 months agowow this is fun!

Mohd Tauheed Khan7 months agoPut x=1

Mohd Tauheed Khan7 months agoI used minimum value of (x+1/x) =2

Amber Deshbhratar11 months ago^{+3}this is a very good USclip channrl

Arvind Kumar11 months agoThe first one has also come in Regional Mathematics Olympiad of India also

Paco Libre11 months agoIf you know paschalโs 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.

etotheipi11 months agoI'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 Lorenzzon11 months agoI tried to add the functions inside each parenthesis on the first question, but it didn't help me out

Quahntasy - Animating Universe11 months ago^{+5}Who does Putnam problem for amusement lol.

LetsSolveMathProblems11 months ago^{+7}You are watching this for amusement, right? =)

ืืืื ื'ืืืืื11 months ago^{+7}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 Mernik10 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 Kim11 months agoI have a request for a problem. Please take a look at the 2006 AIME II (Alternate AIME) #13. Keep up the great videos!

adandap11 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.

LetsSolveMathProblems11 months ago^{+1}Huh, that is strange. It may be a coincidence, but I'm not sure.

el tapa11 months ago^{+1}Oh mate, amazing

77Chester7711 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 Elmutis11 months agoHow comes they are so easy u.u

Sarthak Sharma11 months agoSir 1998 question

Can be simply solved by using AM>=GM inequality

Avishek Tiwari11 months agoUpload more question

Jeff Ahn11 months agoWoow. So elegant.

Siddhartha Shree kaushik11 months ago^{+2}I got 3 notifications within 2 minutes

Are you serious?

Siddhartha Shree kaushik11 months agoPlease stop sending multiple notifications

It's fucking annoying

LetsSolveMathProblems11 months ago^{+1}I did not send multiple notifications. USclip is programmed to issue notifications to subscribers who wish to receive them, and I am not in control of it (as far as I am aware). I apologize for the possible inconvenience.