ques......
1. Find last two digits of ..2003^2004^2005?
2. Find last two digits of 6^11^7 ?
Please Post all answers with full description..................:-P
ques......
1. Find last two digits of ..2003^2004^2005?
2. Find last two digits of 6^11^7 ?
Please Post all answers with full description..................:-P
1)2003^2004^2005mod100
e(100) = 40
2004^2005mod40 = 4^2005mod40 = 2^4010mod8*5
2^4010mod5 = 4
so 8k = 5m + 4
at k = 3
we have 24
so 2003^24mod100 = 3^24mod100 =3^4 * 243^4mod100 = 3^4*43^4mod100 = -19* 01mod100 = -19 = 100-19 = 81
P.S-->here E(100) is the euler's no. of 100
2)6^11^7mod100 = 6^11^7mod5*20
6^11^7mod5 = 1
6^11^7mod20
e(20) = 20(1/2)(4/5) = 8
11^7mod8 = 3^7mod8 = 3
hence
6^3mod20 = 216mod20 = 16
so the no. is of the format: 20k + 16 = 5m +1
at k = 0 we have 16 as tht no.
so the last two digits are 16
ques......
1. Find last two digits of ..2003^2004^2005?
2. Find last two digits of 6^11^7 ?
Please Post all answers with full description..................:-P
2003^2004^2005
as the eulers number for 100 is 40 divide 2004^2005 with 40 and find remainder first
2004^2005 = (2000+4)^2005%40
=4^(16*125)+5 ( eulers of 40 is 16)
= 4^5%40
=1024%40
= 24
Now 2003^2004^2005 = 2003^40k+24 %100
=2003^24
=(2000+3)^24
=3^24%100
=81
last two digits
6^11^7 proceeding in the same way
11^7%40 =11*(121)^3%40
=11
6^11^7 = 6^40k+11
=6^11%100
=56
last two digits
1)2003^2004^2005mod100
e(100) = 40
2004^2005mod40 = 4^2005mod40 = 2^4010mod8*5
2^4010mod5 = 4
so 8k = 5m + 4
at k = 3
we have 24
so 2003^24mod100 = 3^24mod100 =3^4 * 243^4mod100 = 3^4*43^4mod100 = -19* 01mod100 = -19 = 100-19 = 81
P.S-->here E(100) is the euler's no. of 100
2)6^11^7mod100 = 6^11^7mod5*20
6^11^7mod5 = 1
6^11^7mod20
e(20) = 20(1/2)(4/5) = 8
11^7mod8 = 3^7mod8 = 3
hence
6^3mod20 = 216mod20 = 16
so the no. is of the format: 20k + 16 = 5m +1
at k = 0 we have 16 as tht no.
so the last two digits are 16
i think for the second problem u shud not take 20*5 to split 100
coz i guess ..frm chinese remainder theorem in order to split in such way they shud be co primes...
am i correct ??? please enlighten me iam getting 56 as the answer for 2nd one
But ...if we don't go throgh Euler's n go like dat....
1. 2003^2004^2005
^2005
^2005 (when 1 is in the end of digit then multiply ten's digit of dat number to the unit of the exponent to get ten's digit)
^2005
So the Ans is ...41...
2..
dat is nthng...6^77
n cyclicity of 6's ten's digit is 5.
so the reaminder is 2..
so ans will be ...____36.
Please Correct me if m wrong....
But ...if we don't go throgh Euler's n go like dat....
1. 2003^2004^2005
^2005
^2005 (when 1 is in the end of digit then multiply ten's digit of dat number to the unit of the exponent to get ten's digit)
^2005
So the Ans is ...41...
2..
dat is nthng...6^77
n cyclicity of 6's ten's digit is 5.
so the reaminder is 2..
so ans will be ...____36.
Please Correct me if m wrong....
Sorry my frnd u did wrong there.
^2005 => {_81}^xx1
so last two digits 81
and is not 6^77
its (6)^(11^7)
2003^2004^2005
as the eulers number for 100 is 40 divide 2004^2005 with 40 and find remainder first
2004^2005 = (2000+4)^2005%40
=4^(16*125)+5 ( eulers of 40 is 16)
= 4^5%40
=1024%40
= 24
Now 2003^2004^2005 = 2003^40k+24 %100
=2003^24
=(2000+3)^24
=3^24%100
=81
last two digits
6^11^7 proceeding in the same way
11^7%40 =11*(121)^3%40
=11
6^11^7 = 6^40k+11
=6^11%100
=56
last two digits
yea u are rt...i totally forgot tht:)....the ans to the second one shud be 56
Sorry my frnd u did wrong there.
^2005 => {_81}^xx1
so last two digits 81
and is not 6^77
its (6)^(11^7)
I still didn't get that ......
Please explain it again.......:splat:
Q1) Arun,bikas and chetakar have a total of 80 coins among them.arun triples the number of coins with the others by giving them some coins from his own collection.next,bikas repeats the same process.after this bikas now has 20 coins.find the number of coins he had the beginning?
a) 11 b) 10 c) 9 d) 12
Thanks..
is there another option
i'm getting the ans as 20 , please let me know where i'm going wrong
a,b,c have x,y,z coins initially
x+y+z=80.....1
after 1st round
a has=x-2y-2z
b has 3y
c has 3z
after 2nd round
a has 3(x-2y-2z)
b has 3y- 2(x-2y-2z)-6z=20.....2
solving 1 and 2 y=20
I still didn't get that ......
Please explain it again.......:splat:
4th power of 2003 yields 81 as last 2 digits.
So we need to find the units digit of , which is 6 .
multiplying 8 to 6 gives 8 as the 10's digit, so last two digit is 81.
Hope this makes clear.:D
HI guys pls give me detailed approach to solve these
Find t remainder
q1)50^51^52/11
q2)33^34^35/7
How can we apply eulers
wht i hv done is
33^34^35/7
=take 33^34/7
applyng eulers 33^6/7 = 1 (rem)
so 33^34=33^(30+4)/7
= rem 33^4/7=2
now 2^35/7
again apply eulers
we get 2^5/7=rem 4Ans
But the ans is 2
SO PLS LET ME CORRECT
HI guys pls give me detailed approach to solve these
Find t remainder
33^34^35/7
How can we apply eulers
wht i hv done is
33^34^35/7
=take 33^34/7
applyng eulers 33^6/7 = 1 (rem)
so 33^34=33^(30+4)/7
= rem 33^4/7=2
now 2^35/7
again apply eulers
we get 2^5/7=rem 4Ans
But the ans is 2
SO PLS LET ME CORRECT
33^34^37 mod 7
E(7)=6
34 ^37 mod 6
E(6)=2
hence 34^37 mod 6 reduces to remainder of 34/6 that is 4
hence the 34^37 mod 6 is reduced to 6k+4
so the expression reduces to 33^(6k+4) mod 7
33^4 mod7
(35-2)^4 mod 7
(-2)^4 mod 7
16 mod 7 = 2..ANS
Hope this helps...
33^34^37 mod 7
E(7)=6
34 ^37 mod 6
E(6)=2
Note: here u can not apply Euler as 6 and 34 are not co-prime.
hence 34^37 mod 6 reduces to remainder of 34/6 that is 4
hence the 34^37 mod 6 is reduced to 6k+4
so the expression reduces to 33^(6k+4) mod 7
33^4 mod7
(35-2)^4 mod 7
(-2)^4 mod 7
16 mod 7 = 2..ANS
Hope this helps...
Recheck the solutio..it seems to be wrong
tac007 SaysRecheck the solutio..it seems to be wrong
then whts the answer....?
33^34^37 mod 7
E(7)=6
34 ^37 mod 6
E(6)=2
hence 34^37 mod 6 reduces to remainder of 34/6 that is 4
hence the 34^37 mod 6 is reduced to 6k+4
so the expression reduces to 33^(6k+4) mod 7
33^4 mod7
(35-2)^4 mod 7
(-2)^4 mod 7
16 mod 7 = 2..ANS
Hope this helps...
You cant apply euler's there.(bold part)
34^37 mod 6 = (-2)*2^36 mod 6 = -8 mod 6 = -2 or 4.
So, 33^(6k+4) mod 7 = 5^4 mod 7 = 625 mod 7 = 2
hi! im late, but just a last minute doubt daz just not goin outa mah mind..
can you please temme how to find the last two digits of a number, except for the one's anding wid 9 or 1...
P=2^2008 + 1 q=2^2009 +1 r=2^2010 + 1which is prime ?
P=2^2008 + 1 q=2^2009 +1 r=2^2010 + 1which is prime ?
P=2^2008 + 1 and q=2^2009 +1 are not so i guess 2^2010 + 1 is
P=2^2008 + 1 is divisible by 5
and q=2^2009 +1 is divisible by 3
P=2^2008 + 1 q=2^2009 +1 r=2^2010 + 1which is prime ?
2^2009 + 1 = [ (2^7)^287+(1^7)^287 ] = (129)*(...) --->so not prime
2^2010 + 1 = [ (2^670)^3+(1^670)^3 ] = (..) * (...) --- > so not prime
2^2008 + 1 = [ (2^
^251+(1^^251 ] = (257)*(...) --- > so not primea^n + b^n = (a+b)(a^n-1 + ...+ b^n-1) ; (if nis odd ) rule..
So none is a prime
Hello everyone,
I have grt amount of difficulty while i am solving DI section especially bcos of long divison. CAn u pls help me with different methods of divison you are using??
Lets take fr Eg:: 3218/5439
How will be your approach to this???