[nagi_nov3]

Originally Posted by viccy_776

given number is of the form 7+70+700+7000+........7^37

7 when divided by 19 gives remainder 7

7^2 when divided by 19 gives remainder 11


7^3 gives 1 as remainder

again 7^4 is same as 7^3 * 7 which gives remainder 7 again


the cycle repeats itself

7 11 1 7 11 1 and so

2nd last digit gives remainder 1.Hence last digit gives 7

every three remainders add up to 19.hence divisible by 19.only number left will be 7

hence remainder is 7

Hey buddy,

question is not 7^37 mod 19

77777777777.........(37digits)

try gain

[chetna]

Originally Posted by nagi_nov3

Hey buddy,

question is not 7^37 mod 19

77777777777.........(37digits)

try gain

hey nagi whats ur answer

[nagi_nov3]

Originally Posted by chetna

hey nagi whats ur answer

7...37times
=7*10^36+77*10^34+...........+77*10^4+77*10^2+77

77 mod 19 =1

10^2 mod 19 =100 mod 19 =5

so we will left with
7*5^18+5^17+..........+5^2+5^1+1
5^17+..........+5^2+5^1+1 = sum of G.p

7*5^18+ (5^18 -1 )/4

5^18 mod 19 = 6^9 = 216^3 = 7^3 = 343 = 1 mod 19

so we left 7*1+(1-1)/4 = 7

i think a bit long

chetna, post ur solution

[chetna]

right nagi
my approach was just the same....i did a silly mistake on the second last step....

i m improving too

[prof_calculus]

Hi all
I have got some personal messages regarding my attempt at "cow and bridge solution". In addtion to replying to them, I post another message for interested people. The solution posted by me did not get any comment, and subsequently I worked out myself that “the problem does not have a unique solution”.

A cow was standing on a bridge, 5 feet away from the middle of the bridge. suddenly a lightning express with 90 miles/hr was coming towards the bridge from nearest end of the cow. seeing this the cow ran towards the express and managed to escape when the train is one feet away from the bridge. if it would have ran to opposite direction (i.e. away from train) it would have been hit by the train one ft away from the end of the bridge. Calculate the length of bridge.

The solutions:
· Let the length of bridge be 2L and its ends named as A and B with the former one towards the train. The cow is standing L-5 from A and L+5 from B.
· Distance of train from point A when cow saw it be X feet from A
· Speed of Train is 90 miles/per hour or 132 feet per second.
· Cow’s speed is C feet per second.
So we now have two equations:
“managed to escape when the train is one feet away from the bridge” gives us:
Equating Time= Distance/speed --- (L-5)/C =(X-1)/132 ---- (1)
(L+4)/C=(X+2L-1)/132 ---- (2)
Solving it gets us CL=594
It is obvious that the problem can have as many solutions as the factors of 594.
Some of the integral solutions which fulfil all the condition are:
C=66 , L=9 (and 2L=18 length of the bridge) and X=9
C=33 , L= 18 and X= 53
C=22 , L= 27 and X= 132
C=11 , L= 54 and X= 588
My earlier solution was not correct due to some calculation error but the approach was the same.
Anybody trying to fild flaw is welcome , since it would me rectify rectify my approach in future.
_____________________________________

[viccy_776]

Originally Posted by chetna

i didnt get ur solution...
how did u take 7^2,7^3....??
my remiander is 6.......

seems like i was possesed.my fault

[unique_1]

yup.. ans to 7777..(37times) when divided by 19..is 7.
here's my solution.
777...777(37times)=7x(111..111{37times})
now 111..111{37times}=[10^37-1]/9 now taking mod19 of this we have ..
111..111{37times}=1(mod19) [bcoz 10^18=1(mod19)&10^19=10(mod19) by eulers th)..]
hence 777...777(37times)=7(mod19)

[ajay_citm]

Originally Posted by unique_1

yup.. ans to 7777..(37times) when divided by 19..is 7.
here's my solution.
777...777(37times)=7x(111..111{37times})
now 111..111{37times}=[10^37-1]/9 now taking mod19 of this we have ..
111..111{37times}=1(mod19) [bcoz 10^18=1(mod19)&10^19=10(mod19) by eulers th)..]
hence 777...777(37times)=7(mod19)

yaar koi eulers theoram post kar do

[umesh_wpc]

Originally Posted by ajay_citm

yaar koi eulers theoram post kar do

Eular's Theorem:
a^k=1 (mod n)
where
a=any integer not divisible by n
n=any integer
k=number of numbers smaller than n that are coprime to n

formula for calculating k:
1st factorise n=(P1^Q1)(P2^Q2)...(Pr^Qr)
k=n*(1-1/P1)*(1-1/P2)...(1-1/P3)
e.g. n=100
100=(2^2)(5^2)
so k=100*(1-1/2)*(1-1/5)=100*1/2*4/5=40

Example: Find Remainder of 13^256/9
9=3^2
k=9*2/3=6
mod(256,6)=4 (256=42*6+4)
thus mod(13^256/9) => (13^4/9) => mod(169/9)mod(169/9)=mod(7*7)=mod(49/9) =5 ans.

Fremat's Little Theorem:
a^(n-1)=1 (mod n)
also
a^n=a (mod n)
a=any integer not divisible by n
n=prime number

Example: Find remainder for 2000^1000/13?
=>2000^(83*12+4)/13 => mod((2000^12)^83/13).mod(2000^4/13)
=>1*mod(2000^4/13)=(-2)^4/13=16/13
so remainder=3 ans.



Regards
umesh kathuria

[siddharthaduggirala]

HI Suresh,
Can u please solve this one for me!!!
A news paper agent sells TOI,HT and IN in equalnumbers to302 persons. 7 get HTand IN, 12 get TOI and IN 9 get TOI and HT and 3 get all the news papers. Now

1) How many get only one paper
2) What % get TOI or Ht but not IN
3) The difference between the number readinfg HT & IN only and HT only is ??

This is a prob from arun sharma the last one.

Thanks
Siddhartha.