IIT JEE , AIEEE Forum - Mathematics , Algebra

Sushmi

n = 3 ^ 100 then for n

a.) units place is 1 b.) tens digit is 0

c.) units place is 7 d.) tens digit is 2

please dont tell me fermat's theorem and all that please tell me a method using binomial theorem .

LAMPARD

Its a...because,if you see the series
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243...
So,3 raised to the powers of the order 4n have units place as 1.And since 100=4(25),its option a.

Sushmi

Another one of the options is correct.

elastiboysai

Sushmi

No answer given is a and b

Werewolf

write 3¹⁰⁰ as (2-5)¹⁰⁰=ⁿC₀5¹⁰⁰ +[ ^_nC₁2¹5⁹⁹.....and terms having 2 and 5 that is zero in units place] + ⁿC₁₀₀2¹⁰⁰Now 2¹⁰⁰has 6 in units place and ⁿC₀5¹⁰⁰has 5 in units place so we have 5+0+6=11 this means the expression has 1 in units place

LAMPARD

Yes,its b also.
But my answer has got nothing to do with binomial.
See,3^4=81
3^8=6561
3^12=531441
So,as it is evident that the tens digit goes on decreasing as 8,6,4,2,0 and the the cycle repeats again.And since 3^20 has tens place has 0,3^100 also has.
But,we wont be able to calculate powers of 3 in the exam.There is a better method.

sandeepramesh

obvious answer is a and b only.

You can apply chinese remainder theorem to find the last 2 digits of the number itself

sandeepramesh

Note that as 3 and 4 are relatively prime
$3^{\phi(4)} \equiv 3^2 \equiv 1 \mod 4$
$\implies 3^{100} \equiv 1 \mod 4 \implies 3^{100} = 4 \cdot t + 1 \rightarrow$ (1)

Also note that 3 and 25 are relatively prime
Hence $3^{\phi(25)} \equiv 3^{20} \equiv 1 \mod 25$
$\implies 3^{100} \equiv 1 \mod 25 \implies \rightarrow$ (2)

Putting (1) in (2)
$\implies 4 \cdot t + 1 \equiv 1 \mod 25 \implies t \equiv 0 \mod 25$
$\implies$ put $t = 25 \cdot k$
Hence $3^{100} = 100 \cdot k + 1 \equiv 01 \mod 100$
Hence the tens digit is 0 and the units digit is 1 Smile

Sushmi

Sorry sandeepramesh I am not a genius like you, so couldnt understand your process.However thanks everybody.Finally i got it.

sandeepramesh

thas actually not tough to understand.

And to everyone who thought so, i didnt post it to put vetti scene (maybe i did :P) but i posted it as a viable alternative

Decoder

my way..or highway!!..

for unit place try to get (10m +/- 1)^n form expression...
for ten's place..try to get (100m +/- 1)^n form of expression...

if u don't get the latter one..question will have n as a multiple of ten..

write the expression as ( 10 - 1) ^50..

expand ..u get all terms ending wid a zero except last one..i.e 1..

so unit's place is one..

now for ten's place ...
again expand...u will see every term is actually ending wid two zeroes ...except last one..so tens digit is zero...

so correct options r a).b)

nadeemoidu

I don't understand how ppl like sandeepramesh and hsbhatt sir expects everyone to know about modulus and euler theorem.
Here's my answer.

3¹⁰⁰=9⁵⁰

=(10-1)⁵⁰

Expand using binomial theorem ,

= <some multiple of 100> - ⁵⁰C₁10 +⁵⁰C₀1
=<some multiple of 100> -500+1
=<some multiple of 100> +1
so the last two digits are 01.

sandeepramesh

hey, cut the euler theorem part, it was unnecessary .

Also i just gave it for the sake of a soln, bcos i thought it was the shortest to do

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