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General Forum: Education | | ******MATHEMATICS****** | |
| i learned this methods from SDKs Speed Maths Tutor CD there are many more methods in this CD
thanks
Posted by: Mr. Lakshmi Kanth Kethavarapu At: 5, May 2004 12:48:45 PM IST
anyway just kidding!!... 83*87 can b done in the following way... multiply the last 2 digits...21... and since ten's place no is identical.. so multiply the digits with the next greater no... here its 8*9=72 put this no before the previous result .. the answer is 7221.... u can do the same with any number whose one's place has 2 numbers when added gives the sum of 10... and rest of the number before it identical... like
65*65
73*77
91*99
122*128
cheers
Posted by: Mr. Susruth doddapaneni At: 5, May 2004 3:53:12 AM IST
with a calc ofcourse!!!
Posted by: Mr. Susruth doddapaneni At: 5, May 2004 3:49:06 AM IST
wud u mind if i do it in 2 sec?
Posted by: Mr. Susruth doddapaneni At: 5, May 2004 3:38:10 AM IST
Hello everybody,
Does anybody know the method to multiply 83 and 87 in less than 5 seconds.
Posted by: Mr. Lakshmi Kanth Kethavarapu At: 4, May 2004 5:49:19 PM IST
Hi susruth....Nuvvenaa...evaro anukunnaanu....
-ve numbers logic baane undi gaani......actual gaa...positive intezers ani unte oka vela...appudu elaa cheyyaali???
-ve numbers ante,edo question solve cheyyadaaniki saripothundi......but +ve intezers ki chestrhe baaguntundi....
ippudu secretary naa.....evaro mtech vaalle ayyaaru.....sec and joint-sec iddarunu....
Posted by: Raj Sekhar At: 30, Apr 2004 9:03:03 PM IST
the problem set by frank can also b done by the /mod(11) method... only its a bit complicated than wht raj sekhar suggested with 9... but the mod(9) method fails when this case arises,suppose both no are divisible by 9 and the resulting product is something like 4*5963 ..now since the no is divisible by 9 the sum shud also b divisible by 9... so here arises the probability of 9 or 0 getting into the *'s place... i think this can b eliminated by the mod(11) method...
cheers
Posted by: Mr. Susruth doddapaneni At: 30, Apr 2004 8:07:17 PM IST
Raj sekhar chinna doubt, ippudu ikkada telugu association ki student secretary evaru?
Posted by: Mr. Susruth doddapaneni At: 30, Apr 2004 7:45:27 PM IST
raj sekhar ur problem can b solved this way... its the one side array progression type of a problem..
if p^3+q^3+r^3=n where n belongs to Z
and x^2+y^2!=n;
if every -ve integer can b put in p , q , r
then the sum of cubes will also b -ve excepting when x=y=z=0 while any integer which can be put in p , q if squared gives +ve result ofcourse when p=q!=0, so there can b infinite -ve integers which can b put in p,q,r . this cab b refered to the standard theorems in real number theory which state that -ve numbers are infinite. thus infinite solutions can b possible. i dint put this in a real mathematical type of solution format so excuse me for this...
cheers
Posted by: Mr. Susruth doddapaneni At: 30, Apr 2004 7:41:40 PM IST
I too dont know...proceed avvadaaniki koodaa raavatam ledu....thats y posted here...
Posted by: Raj Sekhar At: 27, Apr 2004 9:15:02 PM IST
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