IIT JEE , AIEEE Forum - Mathematics , Algebra - binomial theorem . generic approach for solving these types of problems

cute_sweet_gal

plz mention the generic approach for solving these types of problems(taking below as example)

find the remainder of 5⁹⁹ divided by 13

thnx

shreyasnivas

ok.. general approach is that u express 5^x as some number which differs with a multiple of 13 by a number like 1 or 2..

example 5² = 25 = 26 - 1 = 13 x 2 -1.

so what we do in your question is

multiply and divide by 5.

contd..

shreyasnivas

so we have 5¹⁰⁰ / ( 5 x 13 ).

actually it is smarter to write this as 5 x (5

^{98 x 5}

/ (13))

please omit the step of multiypl n didide.. srry

now 5

⁹⁸

can be written as 25

⁴⁹

this we write as (26 - 1)⁴⁹keep the denominator there only for now.

contd..

shreyasnivas

{[(13 x2) - 1]⁴⁹ x 5 } / 13 is our expression

now we can expand this binomailly

we find that the numerator has 50 terms, and out of them , all except the last term, which is = 1 x 5 is NOT a multiple of 13..

so the remainder is 5.

try doing all this on paper and you will see that it makes sense. reading it here may seem i little scary.

all the best

sandeepramesh

actually this is a logically one step answer with modulos if you know it :)

$5^2 \equiv -1 \mod 13 \implies 5^{98} \equiv -1 \mod 13 \implies 5^{99} \equiv -5 \equiv 8 \mod 13$

So the remainder is 8 Mr. Green Rotfl

Greatdreams

$Good \;work ; however\;it\;can\; be\;written \;as\;:$

$5^{99} = 5^{3}.5^{96} = 125 . (625)^{24}$

$= [13.9 +8](1 + 48.13)^{24}$

$Now \;expand\;binomially\;;$

$Hence \; it\;is=(13.9+8)[1+^{24}C_{1}.48.13 + ^{24}C_2(48.13)^{2}+..................]$

$Hence\; 8\;is\;remainder\;!$

$w.r.t \;the\;previous\;post ;$

$a \;=\;b(mod)c \rightarrow when\;a\;is\;divided\;by\;c\;;b\;is\;the\;remainder$

$P.S-36\;seconds\;differnce\;between\;rabiit\;and\;me.$

$Answer \;has\; been\;edited\;a\;bit!$

sandeepramesh

nice one Greatdreams

sandeepramesh

shreyasnivas man, you must have a lot of patience!

cute_sweet_gal

thnx guys

shreyasnivas

yup.. trying to help.. plus i also didnt think that if help in binomial theorem was reqd.. then it would be pretty difficult to try and understand elementary number thory.. im not too familiar with goldbacks conjecture either..

sandeepramesh

actually elementary number theory is infact elementary Mr. Green Its really easy to understand!

Btw where here is the Goldbach's conjecture used which you are referring to? Mr. Green

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