Problem is easy but what is the answer for this ???
if A=33*32*31*30*29*28
&
if Remainder [ A/(12k)] = 0......then find the minimum value of k
would be " 1 "
would be " 1 "
by wilsons theorm i m gettin answer as 2
is it correct?
Every ceiling, when reached, becomes a floor
38! mod 41
39!mod 41 =1
39*38! mod 41 =1
Let 38!mod41 =x
39x = 41k+1
x= (41k+1)/39 = k +(2k+1)/39
Value of k which satisfies is k =19
So x= 20
=> 38!mod41 =20 (2 is not the answer)
sanjeev-garg Sayswould be " 1 "
How u got the answrr ??? can u explain
orangejawan SaysHow u got the answrr ??? can u explain
33 is divisible by 3 and 28 and 32 are both divisble by 4. So 12 is a factor of the above multiplication.So minimum value of k=1.
38! mod 41
39!mod 41 =1
39*38! mod 41 =1
Let 38!mod41 =x
39x = 41k+1
x= (41k+1)/39 = k +(2k+1)/39
Value of k which satisfies is k =19
So x= 20
=> 38!mod41 =20 (2 is not the answer)
Hello frnd,
Please explain how this step came :
39x = 41k+1
x= (41k+1)/39 = k +(2k+1)/39
********
Hello frnd,
Please explain how this step came :
39x = 41k+1
x= (41k+1)/39 = k +(2k+1)/39
********
We assumed 38!mod 41 =x which when multiplied by 39 will and then divided by 41 will give 1 as the remainder.
So 39x =41k +1
The next step is just to find out values of k which satisfy the equation 😃
rajaramvarun Says33 is divisible by 3 and 28 and 32 are both divisble by 4. So 12 is a factor of the above multiplication.So minimum value of k=1.
This 12K when value taken as 1 will become 121 ie 11*11 >> how come it is divisible ?? Answer is 1
Please explain this step is correct or not
This can be the value or not ?? k =0 also , plss check)
This 12K when value taken as 1 will become 121 ie 11*11 >> how come it is divisible ?? Answer is 1
Please explain this step is correct or not
This can be the value or not ?? k =0 also , plss check)
There is not enough clarity in the question if it is 12*k or 12A. If it is the latter the minimum value of k is 0 as 120 = 4*30 which is a factor of the above multiplication, else the answer is 1.(for the former)
Hello Raja ram, I understood the answer!!! Plss crack this one and let me know ??
What is the general method of solving the question like : find the last three digits of 3^1994 ???
Hello Raja ram, I understood the answer!!! Plss crack this one and let me know ??
What is the general method of solving the question like : find the last three digits of 3^1994 ???
i m getting 049..if dats correct,i'll post the soln..
Hello Raja ram, I understood the answer!!! Plss crack this one and let me know ??
What is the general method of solving the question like : find the last three digits of 3^1994 ???
3^1994mod1000
euler;s no. of 1000 = 64
so 3^10mod1000 = 243^2mod1000 = 049
Hello Raja ram, I understood the answer!!! Plss crack this one and let me know ??
What is the general method of solving the question like : find the last three digits of 3^1994 ???
Last three digits is nothing but finding the remainder when the number is divided by 1000
3^1994 mod 1000= 3^394mod 1000= 3^4 *(3^10)^39 mod 1000
NOw 3^10mod1000 = 049
(3^10)^39 mod 1000 = 49^39 mod 1000 =7^78 mod 1000
Closest multiple of 4 near 78 is 76
7^76 mod 1000
= 2401^19 mod 1000= (2400+1)^19 mod 1000= (1 + 19C1 *2400)mod 1000 = 45601mod1000 =601
So 3^4*7^2*601 mod 1000= 81*49*601 mod 1000 =369
Mani_23 Saysi m getting 099..if dats correct,i'll post the soln..
Answer in Pagalgadha is 369 ?? If u need the solution that I can paste
Mani_23 Saysi m getting 099..if dats correct,i'll post the soln..
i guess the ans is 049...u have typed 9 in place of 4:)
orangejawan SaysAnswer in Pagalgadha is 369 ?? If u need the solution that I can paste
3^1994mod1000
euler;s no. of 1000 = 400
so (3^400)^4*3^394mod1000 =
(3^400)4 will give 1 remainder when divided by 1000..
so we r left with 3^394/1000
orangejawan SaysAnswer in Pagalgadha is 369 ?? If u need the solution that I can paste
I have edited my post.Answer is 369. last 3 digits cannot be obtained by using Euler's directly here and splitting and using Chinese remainder theorem as the numbers thus split also have to be relatively co-prime!
Check my approach.
3^1994mod1000
euler;s no. of 1000 = 64
so 3^10mod1000 = 243^2mod1000 = 049
3^1994mod1000
euler;s no. of 1000 = 64
so 3^10mod1000 = 243^2mod1000 = 049
ryte ans is dis...
049..
(3^64)31 is completely divisible by 1000..
so we r left with 3^10
(3^5)^2/1000
243^2/1000=049..........
Mani and Shanks.Euler of 1000 is 400 and not 64.
E(1000) = 1000 * 1/2*4/5 = 400. (1000 = 2^3 *5^3)
Last three digits is nothing but finding the remainder when the number is divided by 1000
3^1994 mod 1000= 3^394mod 1000= 3^4 *(3^10)^39 mod 1000
NOw 3^10mod1000 = 049
(3^10)^39 mod 1000 = 49^39 mod 1000 =7^78 mod 1000
Closest multiple of 4 near 78 is 76
7^76 mod 1000
= 2401^19 mod 1000= (2400+1)^19 mod 1000= (1 + 19C1 *2400)mod 1000 = 45601mod1000 =601
So 3^4*7^2*601 mod 1000= 81*49*601 mod 1000 =369
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i didn't understood this point :
20k+ 29 = 5c+4
20k+25 =5c
c= 4k+5. or k=1
***********
Can you provide me the document which contains these topics..
>> Every question of finding the last three or two digits can be done like this ???:-P
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i didn't understood this point :
20k+ 29 = 5c+4
20k+25 =5c
c= 4k+5. or k=1
***********
Can you provide me the document which contains these topics..
>> Every question of finding the last three or two digits can be done like this ???:-P
Every question is not so difficult.This is my approach.There might be other approaches too.
Usually CAT level questions will simplify calculations after the use of Eulers. But here since we are left with a significantly high power(i.e 394) we should be able to reduce it to a simpler form.
Mani and Shanks.Euler of 1000 is 400 and not 64.
E(1000) = 1000 * 1/2*4/5 = 400. (1000 = 2^3 *5^3)
euler no 1000=10*10*10
2^3*5^3
euler no=1000(1-1/2)(1-1/4)=400 😃