Harvard University Simplification Tricks | Radical Algebra Aptitude Test | Ivy League Admission Exam

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[@JimS-fs4ub]

Back in the 1950s I learned a technique similar to long division to find square root of any # to any # of decimal places by hand. This problem took less than a minute. I amazed myself that I could still do it.

[@superacademy247]

Wow! 😳

[@w.altran]

Very good. I tried: 11 11 22 22 25 ... 1) 11 is the first pair; and 3x3 and 2 is the rest. 3 is the firt number... 3x2=6; 6__x__ ;The second numer is 3 too. 63x3 =189 ; 211- 189= 22.... 222 22 3 pair is 22... 22222 = 66__x___... the 3... is 3 too... then the 4 number is 3. In the end: 333325 = 6666__x__ ... the last number is 5... I remenber, this is the first method what i learned when I was young!

[@RealQinnMalloryu4]

1^1^1^1^25^1 2^1^1 2^1 (x ➖ 2x+1).

[@crowhunter100]

It is a way to make a simple algebra to a chaotic complication

[@michaeldoerr5810]

The exact answer is (100001)/(3). Looks like I shall use that problem for future practice!!!

[@pedromarcogonzalez6680]

The solution is: I finihsed because I was left without any ink. Seriously, I do not think this way is a simplification trick. Radical, yes, utterly radical.

[@w.altran]

** I remenber, this is the first method I learned when I was young!

[@segaranpillay7154]

every stage on a new page and the last stage cannot be seen .

[@user-jq3ww6qf3s]

35^2=1225 335^2=112225 3335^2=11122225 33335^2=1111222225 333335^2=111112222225 3333335^2=11111122222225

[@superacademy247]

In between 👌

[@YardenVokerol]

a=11111 √1111222225= =√[(a+1)(9a+1)+(2a+3)]= =√[9a²+12a+4]= =√[(3a+2)²]=3a+2=33335

[@mohamadpakzamir]

1111222225=25×59×113×6667=25×6667×6667 Square root 5×6667=33335

[@HenriLaporte-kv6qq]

Sure! And how did you find these numbers?

[@nahumbaskin1883]

11225

[@makneduet]

Madness

[@superacademy247]

That's it. Without calculator 👌🤣

[@user-cn5fs1hg3n]

You have adopted a very long & complicated procedure. How can one know this procedure off hand. Why not to use the normal & well-known method in which we make pairs starting from right side and then we start dividing from left side bringing down pairs every time. If the left-most pair is not complete, it is kept as one digit. Division process is the same which is used in taking square root methods. Why you put people in complicated & difficult procedures.

[@superacademy247]

This method is a warm up for those joining Ivy League Schools