Number System - Questions & Discussions

MBA CAT 2022
2516
testdjjaydev Says
@nitzr7-dude can u plz state d euler's theorem elaborately,,i min wid examples n all..i searched in wikipedia,,but cudnt glean nethng from wat was written dre....thanking u in advance :)

http://www.pagalguy.com/discussions/number-system-25000852
http://www.pagalguy.com/discussions/conceptstotal-fundas-25023536
2517

Find the reminder when +1 is divided by 13

2518
gaurav23march Says
Find the reminder when +1 is divided by 13


My take is 2.

As 13 is a prime number, (13-1)! will leqve remainder by 13 as -1.
So, it is (-1)^14! + 1 = 1+1 = 2

So, remainder is 2.

Note: For any prime number n; (n-1)! leaves remainder as (n-1 which is same -1) and (n-2)! leaves remainder as 1.
2519
Please tell the approach for these kind of problems

Remainder for (2^164)/164 is ??

2^164/2^2*41
2^162/41
(2^10*2^10................16 times)*2*2/41
(1025-1*1025-1*1024-1...................16times)*2*2/41 {41*25=1025}
(-1*-1*-1*-1....................16times)*2*2
1*2*2
4*2^2
4*4
16
2520
gaurav23march Says
Find the reminder when +1 is divided by 13

my ans is......................2
{+1}/13
+1/13
1+1=2
remainder is 2
2521
2^164/164

when power and remainder have some factor common in them take out the common factor
here we have 2^2 common now you are left with 2^162/41
since 41 and 2^162 are co-prime, then2^40/41=1 or 2^160/41=1
so we are left with 2^2*2^2=16 which is the remainder
2522

Originally Posted by Srihari123 View Post
Please tell the approach for these kind of problems

Remainder for (2^164)/164 is ??
2^164/2^2*41
2^162/41
(2^10*2^10................16 times)*2*2/41
(1025-1*1025-1*1024-1...................16times)*2*2/41 {41*25=1025}
(-1*-1*-1*-1....................16times)*2*2
1*2*2
4*2^2
4*4
16

i really dont get it,,neither is the method of cancelling out some common factors correct,,,example....128/96..if the above method is correct then,,we can cancel out the common factors and the answer wud be 2(8/6 left out)..which is absolutely wrong!!
plz help...

2523

what is the remainder when 25! is divided by 10^7? please if you know the ans,replay me with discription

2524

what is the remainder when 128^500 is divided by 153?

2525
ankurba Says
what is the remainder when 25! is divided by 10^7? please if you know the ans,replay me with discription



25!= 5x10x15x20x25 x(1.2.3.4.6.7.8.9.11.12.13.14.3.16.17.18.19.21.22.23.24))
= 5^6 x (2^22)x (1.3.3.7.9.11.3.13.3.17.9.19.21.11.23.3)
= 10^6x 2^16 x3^10 x 7^3x11^2x13x17x19x23

So, dividing it by 10^7 cancels out 10^6 leaving behind just 10

so we have to find out the unit digit (which has to be even)
In this case , the unit digit is 4

so remainder is 4X10^6 =4000000
2526
ankurba Says
what is the remainder when 25! is divided by 10^7? please if you know the ans,replay me with discription


My take is 4000000.

25! has 6 zeros in the end ...(25 => 5+1 = 6)
So, remainder by 10^7 will be last non-zero digit of 25! followed by 6 zeros.

Now, 25 = 5*5 + 0 ....(Putting in form of 5a+b)
Now, R(25) = 2^5 * R(5) * R(1)
= 2^6 = 64 => 4

So, remainder = 4000000.

Note:

ankurba Says
what is the remainder when 128^500 is divided by 153?


My take is 67.

128^500 = 2^(7*500)
153 = 17*9.

Remainder by 17:
2^4 gives remainder by 17 as -1.
So, 2^(4*7*125) will give remainder by 17 as (-1)^(7*125) = -1

Remainder by 9:
2^6 gives remainder by 9 as 1.
Now, 2^(7*500) = 2^3500 = 2^(6k+2)
So, remainder by 9 for 2^3500 = 2^2 = 4

So, we need to find a number of form 9x+4 = 17y-1 => 67
2527

@ankurba

Make sure your thorough with Chinese remainder theorem,Fermat's little theorem ,Euler number and Carmichael number


@ThinkAce ..thank you

2528

25^2-20^2=225
17^2-8^2=225
113^2-112^2=225
Ans 3

2529
ankurba Says
what is the remainder when 128^500 is divided by 153?

16 ans hai kya
2530

puys....can anybdy provide CRT in a simple lingo.....

2531
montupatel_05 Says
puys....can anybdy provide CRT in a simple lingo.....


try this
http://www.pagalguy.com/discussions/number-system-25000852


post number 1858
2532

25^102/17. euler's number of 17 is 17(1-1/17)=16. 102/16 remainder is 6. 25/17 remainder is 8. this sum can be reduced to 8^6/17= 2^18/17= 2^18/(16-(-1)) =(4) (2^16/(16-(-1))) (4) ((2^4)^4)/(16-(-1))
this can be taken further by remainder theorem . the answer for 1st question is 4. similarly the other question can also be solved using this method.

2533

Originally Posted by ankurba
what is the remainder when 128^500 is divided by 153?


Is the answer 7 ???
2534
25^2-20^2=225
17^2-8^2=225
113^2-112^2=225
Ans 3


correct ans 5
2535

can anybody provide the internet link for euler and chinese method?