hi all,
could anybody throw some light on the Euler method of solving to get remainders? i went thru a couple of quant books, but it is nowhere to be found!! looking forward to hearing from you!!
thanx
99.......99 (2009 times) = n
whats the sum of digits of n^2
9999----------------------2009 times can be written as (10^3000-1)
n^2=(10^2009-1)^2=
10^2009+1-2*10^2009
=10^4018+1-20^2009
=9*2008+8+1=18081
another way 99*99= 9801
999*999=998001
9999*9999=99980001
if we see the pattern with 4 9's * 4 9's product we r getting contains 3 9's and 1 8 and 3 zeros
similarly for 9999---2009 times we get 9*2008 +8+1=18081
99.......99 (2009 times) = n
whats the sum of digits of n^2
is the answer 18081
99.......99 (2009 times) = n
whats the sum of digits of n^2
answer should be 9*2009 = 18081
99.......99 (2009 times) = n
whats the sum of digits of n^2
Is the answer 18090?
naga25french Saysanswer should be 9*2009 = 18081
I guess it is the correct answer
as 999....n times ^2
follows the following pattern
(n-1) times 9 followed by 1 "8" followed by zero and the last digit as 1
Hence 2002*9+8+1
=2009*9
as suggested by Naga
I guess it is the correct answer
as 999....n times ^2
follows the following pattern
(n-1) times 9 followed by 1 "8" followed by zero and the last digit as 1
Hence 2002*9+8+1
=2009*9
as suggested by Naga
answer is 18081
99^2=9801
so (99...2009 9s)^2 will be of form 9999(2008 times)8(00000(2008 times)1
sum =9*2008+8+1
18081.....:) other approaches were also good
hi all,
could anybody throw some light on the Euler method of solving to get remainders? i went thru a couple of quant books, but it is nowhere to be found!! looking forward to hearing from you!!
thanx
hey i got this from somewhere(might be from some thread)
see if it hlps
Co-Primes are very important , when it comes to finding remainders.
First things first, What is a Co-Prime ?
2 numbers are said to be co-primes/relatively prime if their HCF is 1.
That is the 2 numbers have no other common factor apart from the number.
Co-Primes by themselves serve no purpose.You need to use them with
Fermat's theorem / Chinese Remainder Theorem etc.
Fermat's theorem states that
Remainder (x ^ ( Euler's Number of N) )/N ) = 1
Here X and N have to Co-Primes.
Now , what is Euler's Number?
Euler's number of a particular number is the number of co-primes less than
the number. The total number of co-primes that a certain number can have
is infinite. So we need to determine the number of coprimes less than the number.
Euler's number for a prime number P is P-1. As all numbers less than that
prime number will coprimes of the prime number. As in 1,2,3,4,5,6 will all be
coprimes of 7.
For other numbers :
Euler's Number of a number X = X(1-1/a)(1-1/b)....
Here a and b are the prime factors of the number.
We look at the number of the prime numbers that are factors of the given number.
We ignore the number of times the prime number occurs.
So say the number is 18.
18 has 2 and 3 as factors.
So Eulers No of 18 = 18(1-1/2)(1-1/3)
= 6
So if we get a sum like ,
Find the remainder when (11 ^ 97 )/7
1)See if 11 and 7 are relatively prime to each other
2)If they are relatively prime, Find the Euler's number of the denominator.
3)Express power in terms of the euler's number.
For eg, in this case the Euler's number for 7 is 6.
So 97 = 6k+1.
We know that if we can factorize the numerator in a division, the end remainder
will be the product of the remainders of each individual factor.
So this becomes (11 ^ 6)/7.....(16 times)* (11 ^ 1)/7
Now as per Fermat's theorem,
(11 ^ 6) /7 is 1.
So the answer is the remainder when (11^1) is divided by 7.
This can be used when the numerator can be factorized.
This can be used when the numerator and the denominator are co-primes/relatively prime.
Now The question arises what is they are not coprimes.
If 2 numbers are not co-primes,it means they have a common factor.
See if you can remove the common factor from both the numerator and denominator.
In case that can't be done , we will have to use other methods to find the remainder.
hey guys one more from simcat
unit digit of (323)^x where x=(7AB847) base 16 is
naga25french Saysanswer should be 9*2009 = 18081
Please share your approach naga.
hey guys one more from simcat
unit digit of (323)^x where x=(7AB847) base 16 is
(323)^x = (323)^(4k+7)
unit digit = 7
hey guys one more from simcat
unit digit of (323)^x where x=(7AB847) base 16 is
It is of base 16 hence that will be a multiple of 4 since last digit is 7
It is of the form 4k+3
Hence the remainder will be
7
(323)^x = (323)^(4k+7)
unit digit = 7
can you explain how to convert base 16 to decimal and vice versa
sachy24 Sayscan you explain how to convert base 16 to decimal and vice versa
x=(7AB847) base 16
in decimal system:-
7 + (4*16^1 + 8*16^2+......) = 4k+7 = 4p+3
Hi puys,
can any1 solve the problem in a lucid way?
N=888888888...89 digits % 470=?
less than one to be a champion
b.302
How come it is 303-1 = 302.
If it is Round Robin league fashion then ans is what?
BTW how many TYPES of playing techniques are there?
How come it is 303-1 = 302.
If it is Round Robin league fashion then ans is what?
BTW how many TYPES of playing techniques are there?
The question says the matches are singles elimination matches. So, in the first round 1 team will play once and the second team is eliminated...
Hi puys,
can any1 solve the problem in a lucid way?
N=888888888...89 digits % 470=?
is the answer 248
Hi puys,
can any1 solve the problem in a lucid way?
N=888888888...89 digits % 470=?
470=10*47
8*10^88+8*10^87+8^86+-------------------------------------8
8(10^89-1)/9 mod 47
8(10^89-1)/9 =k
8*10^89=9k+8
(8*10^92)mod 47=1000(9k+
mod 479000k mod 47 =2
k=43
47 x+43=10y+8
y=47x+35/10
x=5,y=27
so remainder is 278
Find the value of 1/ (1+1/ (3-4/ (2+1/ (3-1/2)))) + 3/ (3-4/ (3+ (1/ (2-1/2))))
a. 13/7 b. 15/7 c. 11/21 d. 17/28