Number System

[shashank3012]

@shivam,
how 888...ending with 8 zeroes,divisible by something that ends with 10 zeroes.
Also,could you please explain how 888...880times is divisible by 45?
Thanks.

[shivam_01]

Originally Posted by masoom (Number System)
can u find reminder for 888888888888......(888 times) by 45!

@ Nidhi and @Shashank

I am not completely sure but still posting my answers

We have to find 888888888888......(888 times) by 45!

so finding the number of zeros in the 45! that is
[45/5]+[45/25]= 10

so we can write 888888888888......(888 times) as
= 888888888888......(878 times)*10^10 + 8888..(10times)

as we can write 8888 as 8800+88

so in this case first term would be divisible and hence only the second term will be needeed the remainder

so 8888..(10 times) mod 45!
i tried using the individual division method and the same remainder is coming

8888..(10 times) as the answer

@ Nidhi there is no as rule for finding the remainder in case there is factoraila in the denominator so i just tried to break it up and found the remiander by individual methods...

correct me if i am wrong

[shashank3012]

Shivam,i dint understand that step where you said that,
888....880 times*10^8 is divisible by 45!.Is there any shortcut to find it.
Because,45! Has 10 zeroes and the numerator is with 8 zeroes.So,two zeroes remain.
Rest is correct.

[shivam_01]

Originally Posted by shashank3012

Shivam,i dint understand that step where you said that,
888....880 times*10^8 is divisible by 45!.Is there any shortcut to find it.
Because,45! Has 10 zeroes and the numerator is with 8 zeroes.So,two zeroes remain.
Rest is correct.

Haan yaar correct kar diya
pata nahi kya sooch raha that jo aisi galti kari

method sahi lagaya tha par typo error kar di......

[peeyushmehta99]

Originally Posted by gaurishankar

how did u got the last two digits of 81^413 so eaisly
plz. explain.......

plz help me to find last 2 digits of-
1). 101*102*103*197*198*199
2). (201*202*203*204*246*247*248*249)

[peeyushmehta99]

plz help me to find last 2 digits of-
1). 101*102*103*197*198*199
2). (201*202*203*204*246*247*248*249)

[jain12345]

Originally Posted by peeyushmehta99

plz help me to find last 2 digits of-
1). 101*102*103*197*198*199
2). (201*202*203*204*246*247*248*249)

answer for the 1st is 64
its very easy
find out the remainder when the number (101*102*103*197*198*199) is divided by hundered
so finally it somces out to be 1 x 2 x 3 x -3 x -2 x -1 which is equal to -36
thus positive remainder is 64..
so answer is 64

try and solve the 2nd one..

cheers

[shivam_01]

Originally Posted by peeyushmehta99

plz help me to find last 2 digits of-
1). 101*102*103*197*198*199
2). (201*202*203*204*246*247*248*249)

just take the last digit of each number and keep on multiplying and the remainder comes out to be 64
or divide each term individually by 100 and multiply the remainder answer 64

same for the second one answer is 76

[peeyushmehta99]

thankzz buddy

[Stella@nokia]

i guess its 72...but m sure this not the correct ans....coz we have 1000 as denom...so ans will b in decimal plz tell the correct ans[/quote]


Hi guys .. am sorry for the late reply ...couldnot get access to internet!! newaz ..
The ques is (402^257)/1000 as asked by ppl..
But unfortunately i dont have an answer to this ques .. i just thought of this expression and tried my hande on it .. Can u plz explain the method u used to solve it ..
there is another similar ques from BYJU's Cat classes for which i have the options but still the soln is not available.
The ques is " Find the remainder of (402^12)/1000? "
options: (a) 496 (b)696 (c) 896 (d) None of these

Plz share the method followed . I tried to apply both fermat's theorm and power cycle bt unsuccessful ...

thanks