I posted this at the end of the Test #2 thread, but I decided it might need it's own thread...sorry for being redundant-redundant...hindsight 20/20 speech another time...
As for #57, this is one of my favorite NS shortcuts because the algebra is so cool...when this question first came to my attention (I used to say when it was first put on a NS test, but I quickly learned that was almost always an inaccurate statement to use concerning NS...what goes around will most certainly eventually come back around...), we had to wait 3 long months for it to reappear on the 1st test of the year after being on the State test the previous year...not enough of a pattern to understand what was happening...we spent the summer guessing (mostly incorrectly)...of course, that is the fun part, but I digress...
Let this problem be modeled for a second by (AB)^2 + (CD)^2, where A, B, C & D represent digits (for example, in 36^2 + 57^2, A=3, B=6, C=5 & D=7)
To fit the structure of this problem, A+D=10 and B=C+1 (B and C are descending).
Now, to describe the algebraic model for this problem, let's refer the the standard model of a 2-digit number, 10x+y...using 10x+y for the 1st number (AB), the entire problem would be set up like this:
So, 36^2 + 57^2 = 101(3^2 + 6^2) = 101(9+36)= 4545..the 2nd number, 57, in the problem becomes inmaterial to the calculation process.
As for #57 on Test 2, 54^2 + 66^2. To get it in the proper form, switch the numbers being squared, which means that 101(6^2 +6^2) = 7272
Boy, that was sure cooler to do on a chalkboard in front of a bunch of "oohing and aahhing" NS players than "picking and grinning" alone behind the keyboard...hopefully it made sense because it truly is one of my favorite tricks...(nothing like diving right in the deep end with your first posting, huh???...)
BTW, my team and I just got word of this Summer League and we are playing catch-up as quickly as possible...I took both tests this evening and I have my team taking them tonight and tomorrow...I'll e-mail all of our scores hopefully tomorrow to get us caught up...this sounds fun, even though Newberry was already at my house tonight talking smack...gonna be a long summer in the oldtimer division...Get him Aaron, get him!!!...
You can only use this method when A+D = 10 and | B-C | = 1....
So you take the digits of the number whose inner digits are larger ie. when arranged like AB CD, B and C are the two youre comparing.... and then you take the digits of the number that was larger, and multiply the sum of each digit squared by 101.
So by example I have:
45^2 + 46^2 = _________
we take the number 56 b/c 5>4....
so we set up the equation:
101(4^2 + 5^2) = 101(16 + 25) =101(41) = 4141
I tink' dats right.
anyway...
Thanks for the cool explanation.
Send yalls scores in as soon as possible.
ps. Aaron is on top. He beat Mr. Newberry by 9 points on the first test...
I think you've got it...almost...assuming you meant absolute value of B-C=1...it has to be B-C=1...B must be bigger, otherwise you would get:
[10y+10+10-x]^2, which won't be pretty at all...
here are 4 easier to see examples...once you recognize them, then when they reverse the numbers, you can still spot them...don't worry, the first time they reversed the numbers, we though it was a new trick...forest for the trees issues!!...
39^2 + 87^2...3+7=10 & 9-8=1, so 101(3^2 + 9^2)=101(90), so 9090
57^2 + 65^2...25 + 49 = 74, so 7474
61^2 + 4^2...36 + 1 = 3737
96^2 + 51^2 = 101(81+36) = 101(117) = 11817
Only time it gets tough is when the two digits you square are so big that their sum is over 100... hope that clears up my mud from earlier...
