Question

5^99 divided by 13 what is the reminder ?

Answer 1

Before I go through the solution, let me ask first, are you aware about modulo or modulus method?

Answer 2

nope
can you answer this on the basis of binomial theorems?

Answer 3

.718420321

Answer 4

8

Answer 5

Well, let me try it first, it has been a while I have tried this method. Give me some minutes

Answer 6

7 or 8 depending on to what place you have toround

Answer 7

@OwlGirl212 how did u get that?

Answer 8

Working mod 13, note that 5^(13-1) = 1 (mod 13).

So, 5^99 = 5^(12 * 8 + 3)
.............= (5^12)^8 * 5^3
.............= 1^8 * 125
.............= 8 (mod 13).

Since 8 is between 0 and 12 inclusive, the remainder in question equals 8.

Answer 9

@OwlGirl212 - The solution you posted doesn't belong to you! It is considered as Plagiarism. Quoting some other person's work without stating the owner or writer of that part is considered as Plagiarism.

Please prefer to post your own solution in future.
If, due to any reasons you have to copy-paste a solution from other resource, then do mention the link

Here is from where you copied it : https://in.answers.yahoo.com/question/index?qid=20120521102952AA9NyEn

Answer 10

@ganeshie8 - Can you help him? I am not sure whether I will be able to help him ... I have to leave now. Sorry!

Answer 11

someone please help me ..

Answer 12

\(\large (x+y)^n = \sum \limits_{k=0}^n \binom{n}{k} x^{n-k}y^k\)

Answer 13

\(\large 5^{99} = 5\times 5^{98} = 5\times (5^2)^{49} = (26-1)^{49} \)

Answer 14

Notice that 26 leaves 0 as remainder when divided by 13.

Answer 15

think wat terms u wud get when u expand above^

Answer 16

\(\large (26-1)^{49} = \sum \limits_{k=0}^{49} \binom{49}{k} 26^{49-k}(-1)^k \)

Answer 17

\(5^{99} = 5\times 5^{98} = 5\times (5^2)^{49} \\ = 5 \times (26-1)^{49} = 5 \times \left( 26(somthing) -1\right)\)

Answer 18

divide it by 13,
whats the remaider ?

Answer 19

divide what?

Answer 20

\(\large \frac{5 \times \left( 26(somthing) -1\right)}{13} \)

Answer 21

but how to divide that eq?

Answer 22

easy, keep in mind what you're after

Answer 23

\(\large \frac{5 \times \left( 26(somthing) -1\right)}{13} \)


\(\large \frac{5 \times 26(somthing) -5}{13} \)

Answer 24

since 13 divides 26 fully, any multiple of 26 wud leave 0 as remainder

Answer 25

right ?

Answer 26

ok

Answer 27

\(\large \frac{5 \times 26(somthing) -5}{13} \)


\(\large \frac{5 \times 26(somthing) }{13} - \large \frac{5}{13} \)


\(\large 0 - \large \frac{5}{13} \)

Answer 28

^thats the remainder

Answer 29

-5 or -5+13 = 8 both are same

Answer 30

how did that +13 come?

Answer 31

you can add 13's when u divide by 13

Answer 32

adding 13's will not affect ur remainder

Answer 33

13 is same as 26 is same as 39 is same as -13 when u talk about remainders when divided by 13

Answer 34

so if i get a negative remainder i need to add it with the divisor??

Answer 35

5^12=1 mod 13
5^(12*8)=1^8 mod 13
5^96=1 mod 13
5^3=8 mod 13
5^99=8 mod 13

Answer 36

if u want a positive answer, yes

Answer 37

saying the remainder is "-5" is perfectly fine as well

Answer 38

but in general ppl like positive numbers as remainders, so... add 13

Answer 39

@ganeshie8 - this is probably a silly question but I can't see how you got
5*(26(something - 1)

Answer 40

just that last step

Answer 41

oh.. sorry i dint show some steps in between... one sec... :)

Answer 42

\(\large 5^{99} = 5\times (26-1)^{49} = 5\sum \limits_{k=0}^{49} \binom{49}{k} 26^{49-k}(-1)^k \)


\(= 5 \left[ \binom{49}{0} 26^{49-0}(-1)^0 + \binom{49}{1} 26^{49-1}(-1)^1 + ... +\binom{49}{49} 26^{49-49}(-1)^{49} \right] \)



\( =5 \left[26(somthing) + \binom{49}{0} 26^{49-49}(-1)^{49} \right] \)

Answer 43

in the binomial expansion,
all terms except the last term has a factor 26 in common... so we can pull it out

Answer 44

corrected the typo :


\(\large 5^{99} = 5\times (26-1)^{49} = 5\sum \limits_{k=0}^{49} \binom{49}{k} 26^{49-k}(-1)^k \)


\(= 5 \left[ \binom{49}{0} 26^{49-0}(-1)^0 + \binom{49}{1} 26^{49-1}(-1)^1 + ... +\binom{49}{49} 26^{49-49}(-1)^{49} \right] \)



\( =5 \left[26(somthing) + \binom{49}{\color{red}{49}} 26^{49-49}(-1)^{49} \right] \)

Answer 45

ah i see it now thanks

Answer 46

np :)

Answer 47

thank u so much guys