prats92 Saysdid not understand how tat 13^2 become -1 plz could u explain tat :grin:
13^2/17= 169/17
so remainder is -1 ( negative remainder concept)
13^2/17= 169/17
so remainder is -1 ( negative remainder concept)
find the remainder when (38^16!)^1777 is divided by 17
ans: E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1
I dont understand this. I googled a lot about FermatEuler theorem but i dont seem to understand them. Somebody please assis me on how do we arrive at this answer. And please throw some light on FermatEuler theorem.
find the remainder when (38^16!)^1777 is divided by 17
ans: E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1
I dont understand this. I googled a lot about FermatEuler theorem but i dont seem to understand them. Somebody please assis me on how do we arrive at this answer. And please throw some light on FermatEuler theorem.
how to find mod of a number...
suppose you have to find mod of 17
17.(1-1/17)=17.16/17=16
you just have to find prime factors of the number...and then multiply that number by (1-1/x1)(1-1/x2)....
where x1,x2...are prime factors.......
so here...mod 17=16
how to find mod of a number...
suppose you have to find mod of 17
17.(1-1/17)=17.16/17=16
you just have to find prime factors of the number...and then multiply that number by (1-1/x1)(1-1/x2)....
where x1,x2...are prime factors.......
so here...mod 17=16
thank you so much!!! 😃
(36^41)/7*11
(36^41) mod 7 = 1
(36^41) mod 11 = 3
so remainder
7x+1 = 11y+3
x=(11y+2)/7
put y = 3
x = 5
so (36^41) mod 77 = 36
(41^36) mod 7*11
(41^36) mod 7 = 1
(41^36) mod 11 = 3
so from here also remainder = 36
now 2*36 mod 77 = 72..correct me if i am wrong
(36^41) mod 7 = 1
(36^41) mod 11 = 3 How did you find the reminders for these?? Please help. Im struggling with this concept.
so remainder
7x+1 = 11y+3
x=(11y+2)/7
put y = 3
x = 5 Also help me with this too. I dont get the logic.
(36^41) mod 7 = 1
(36^41) mod 11 = 3 How did you find the reminders for these?? Please help. Im struggling with this concept.
so remainder
7x+1 = 11y+3
x=(11y+2)/7
put y = 3
x = 5 Also help me with this too. I dont get the logic.
With the eulers theory the first equation reduces to 36^5 mod 7 , which is 1
similarly the second eq reduces to 36^1 mod 11 = 3.
from remainders theorm,
the 1st eq when divided by 7 gives remainder 1 so 7x+1 .
similarly 11y+3.
hope it helps
With the eulers theory the first equation reduces to 36^5 mod 7 , which is 1
similarly the second eq reduces to 36^1 mod 11 = 3.
from remainders theorm,
the 1st eq when divided by 7 gives remainder 1 so 7x+1 .
similarly 11y+3.
hope it helps
I understood that the equation reduces to 36^5 mod 7 by eulers theorem. But how did you find the reminder to be one after that. Is it by finding out 36^5? Or is there any simple method?
acidguy23 SaysI understood that the equation reduces to 36^5 mod 7 by eulers theorem. But how did you find the reminder to be one after that. Is it by finding out 36^5? Or is there any simple method?
36^5/7= 1
yaar,here simply divide 36 by 7, you will get 1^5/7= 1
the last two digits of the given expression are :
101*102*103*197*198*199
2)65*29*37*63*71*87
the last two digits of the given expression are :
101*102*103*197*198*199
2)65*29*37*63*71*87
Since, 100 = 4*25
1) N = 101*102*103*197*198*199 = 1*2*3*(-3)*(-2)*(-1) (mod 25) = 14 (mod 25)
N = 101*102*103*197*198*199 = 0 (mod 4)
=> N = 25k + 14 = 4n
=> n = 6k + 3 + (k + 2)/4, n will be integer for k = 2
=> N = 100m + 64
So, last two digits will be 64
2) N = 65*29*37*63*71*87 = 1*1*1*3*3*3 (mod 4) = 3 (mod 4)
N = 65*29*37*63*71*87 = 15*4*12*13*21*12 (mod 25) = 60*156*252 (mod 25) = 10*6*2 (mod 25) = 20 (mod 25)
=> N = 4n + 1 = 25k + 20
=> N = 100m + 95
=> Last two digits will be 95
the last two digits of the given expression are :
101*102*103*197*198*199
2)65*29*37*63*71*87
for finding last two digits divide by 100
1) 1*2*3*97*98*99
=1*2*3*(-3)*(-2)*(-1)
=-36
=64
2)35*29*37*37*29*13
=1015*1369*377
=15*69*77
=79695
=95
hey i am new to pagal guy and preparing for cat 2011. i am not able to understand the above posts on eulers theorem.how do i use it to find the remainder..i read the above replies but i am still not clear..please help me wid dis..
Since, 100 = 4*25
1) N = 101*102*103*197*198*199 = 1*2*3*(-3)*(-2)*(-1) (mod 25) = 14 (mod 25)
N = 101*102*103*197*198*199 = 0 (mod 4)
=> N = 25k + 14 = 4n
=> n = 6k + 3 + (k + 2)/4, n will be integer for k = 2
=> N = 100m + 64
So, last two digits will be 64
2) N = 65*29*37*63*71*87 = 1*1*1*3*3*3 (mod 4) = 3 (mod 4)
N = 65*29*37*63*71*87 = 15*4*12*13*21*12 (mod 25) = 60*156*252 (mod 25) = 10*6*2 (mod 25) = 20 (mod 25)
=> N = 4n + 1 = 25k + 20
=> N = 100m + 95
=> Last two digits will be 95
why was the number 100 split into 2 parts 4 and 25???
dntcheatme Sayswhy was the number 100 split into 2 parts 4 and 25???
In case of question, there was no need to do so, but in 2nd question its easier to find the remainder by 25.
Generally its easier to find the remainder for a smaller number.
but while writing a number N as a*b, remember that a and b should be coprime.
Hi fellow puys!
Here is an easy but interesting one for all of you:
Find the sum of the digits of the smallest integer whose first digit is 1 and which has the property that if this digit is transferred to the end of the number the number is tripled.
a) 22
b) 27
c) 36
d) 37
e) 39
Hi fellow puys!
Here is an easy but interesting one for all of you:
Find the sum of the digits of the smallest integer whose first digit is 1 and which has the property that if this digit is transferred to the end of the number the number is tripled.
a) 22
b) 27
c) 36
d) 37
e) 39
Let the number be (10^n) + b
=> 3(10^n) + 3b = 10b + 1
=> b = (3*10^n - 1)/7
Smallest value of n for which b comes out to be an integer is 5
=> b = 42857
=> Number is 142857
=> Sum of digits = 27
Hi fellow puys!
Here is an easy but interesting one for all of you:
Find the sum of the digits of the smallest integer whose first digit is 1 and which has the property that if this digit is transferred to the end of the number the number is tripled.
a) 22
b) 27
c) 36
d) 37
e) 39
option b..!!
use more of a logic.. here to solve..
as when 1 is transferred at the end then no. becomes 3 times the original no. therefore the number at the end must be 7 as it yields 1 when multiply it with 3, also the remaining digit must be the same as original number therefore next number must be 7 from the end , and we obtain 7 by multiplying 3 with 5 as 2 was carry obtain by multiplying 3 with 7.. in a similar fashion we can obtain the number as 142857..!!
With the eulers theory the first equation reduces to 36^5 mod 7 , which is 1
similarly the second eq reduces to 36^1 mod 11 = 3.
from remainders theorm,
the 1st eq when divided by 7 gives remainder 1 so 7x+1 .
similarly 11y+3.
hope it helps
i am stil confused wid the eulers concept..how did the equation reduce to 36^5 mod 7. sry for being a little dumb..:-P
dntcheatme Saysi am stil confused wid the eulers concept..how did the equation reduce to 36^5 mod 7. sry for being a little dumb..:-P
36/7 (Mod) = +1
therefore (+1)^5 = +1 hence remainder is +1. You don't require Euler's number here.
But the next case for (36^41) mod 11 = 3, you need Euler's Number.
Euler's Number :
Euler's number of N = a^x * b^y... where a, b,.. are prime numbers
is given by E(N) = N * * ...
Hence Euler's number for 6 = 2 *3 is 6 * (1/2) * (2/3) = 2.
Now fermat's Little theorem states that for prime number, Euler's number is (prime number -1), hence for 11, euler's number is 10, likewise.
Now why we require Euler's Number.
We require Euler's Number as {X^}/N (Remainder) = +1 (Provided X and N are COPRIMES)
hence E(6) = 2 (as above)
And 5^2/6 (Remainder) = 25/6 (Mod) = 1
Now for the above problem (36^41) mod 11 = 3
Rules: 1) Reduce the base by the divisor
2) Reduce the power by the E(Divisor)
36/11 (Mod) = +3
41/10 (Mod) = +1
hence (36^41) mod 11 = 3^2 mod 11 = +3
I hope the concept is clear.
36/7 (Mod) = +1
therefore (+1)^5 = +1 hence remainder is +1. You don't require Euler's number here.
But the next case for (36^41) mod 11 = 3, you need Euler's Number.
Euler's Number :
Euler's number of N = a^x * b^y... where a, b,.. are prime numbers
is given by E(N) = N * * ...
Hence Euler's number for 6 = 2 *3 is 6 * (1/2) * (2/3) = 2.
Now fermat's Little theorem states that for prime number, Euler's number is (prime number -1), hence for 11, euler's number is 10, likewise.
Now why we require Euler's Number.
We require Euler's Number as {X^}/N (Remainder) = +1 (Provided X and N are COPRIMES)
hence E(6) = 2 (as above)
And 5^2/6 (Remainder) = 25/6 (Mod) = 1
Now for the above problem (36^41) mod 11 = 3
Rules: 1) Reduce the base by the divisor
2) Reduce the power by the E(Divisor)
36/11 (Mod) = +3
41/10 (Mod) = +1
hence (36^41) mod 11 = 3^2 mod 11 = +3
I hope the concept is clear.
hey thank you so much..the concept is really getting into my head now..this means that fermets theorem is a special case of the above stated theorem which says /p (where p is a prime number) has a remainder equal to 1.
so eulers concept can be applied to every other problem of remainders involving big values????