ankurba Sayswhat is the remainder when 25! is divided by 10^7?
last non zero digit of 25! is 2 so remainder will be 2 because 25! has 6 zero's
last non zero digit of 25! is 2 so remainder will be 2 because 25! has 6 zero's
no no when u take common after finding the remainder u have to multiply that common with remainder
i guess
Hi Puys,
Can anyone pls explain how to solve below problem by Chinese theorem:
Whats the remainder when 123412341234......(upto 400 digits) is divided by 909.
thanks,
Gaurav K
Look how i approach
1234 =10 sum of digits
so that 10*100=1000
now 909=9*101
when u divide the 1000 with 9 u will get 1 remainder
by using chinese remainder theorem now
u llook into the option there is only one option that gives u 1 remainder by divinding by 9 so that is the anwesr
i dont remember the answer
but this is how i approach
to learn more about chinese remainder theorem see the RAVI HANDA SIR's posts
u Will find it .
Hi Puys,
Can anyone pls explain how to solve below problem by Chinese theorem:
Whats the remainder when 123412341234......(upto 400 digits) is divided by 909.
thanks,
Gaurav K
1234...400 digits
909 = 9*101
100*(1+2+3+4) = 100*10
R(1000)/9 = 1
now say you have more than 1 options which are of the form 9x+1
check for 101.
how?
the divisibility for 101 is take last 2 digits from the right and subtract next 2 digits
i.e. (34-12)*100
= 22*100/101 = -22
so number should be of the form
9x+1 = 101y+89
i'm also getting 96 days..
84 *365=30660
less-84*313=26292
4368 days +21 days as leap one=4389
4389/365=12 appx
84+12=96 yrs...
how did ya people getting 98 yrsplz explain lucidly
plz explain lucidlyhow can we say tat all 21 days of d leap year part will b sunday only??? it can be odr day also. still takin that part into account we get d answer 2 b approx 96 years.
Hi Puys,
Can anyone pls explain how to solve below problem by Chinese theorem:
Whats the remainder when 123412341234......(upto 400 digits) is divided by 909.
thanks,
Gaurav K
909=9*101
(1+2+3+4)*100=1 (mod9)
for 101, (32-12)*100 = 79 (mod101)
so remainder will be of the form
9m+1=101n+79
685???
Hi Puys,
Can anyone pls explain how to solve below problem by Chinese theorem:
Whats the remainder when 123412341234......(upto 400 digits) is divided by 909.
thanks,
Gaurav K
123412341234...(400 digits) mod 9 = 1
so Rem. mod 9 =1
only one answer satisfied in the mock i.e. 685
ankurba Sayswhat is the remainder when 128^500 is divided by 153?
153 = 17*9
E(9)=6
500 mod 6= 2
128 mod 9= 2
=>128^500 = 4 (mod9)
E(17)=16
500 mod 16= 4
128 mod 17= 9
128^500 = 9^4 (mod17)
=> 9^2*9^2 mod 17 = 13*13 mod 17 = 169 mod 17 = -1
so rem will be of the form 9m+4=17n-1
909=9*101
(1+2+3+4)*100=1 (mod9)
for 101, (32-12)*100 = 79 (mod101)
so remainder will be of the form
9m+1=101n+79
685???
Yes Enceladus...thanks....
Also thanks chandrakant and barclay 😃
1234...400 digits
909 = 9*101
100*(1+2+3+4) = 100*10
R(1000)/9 = 1
now say you have more than 1 options which are of the form 9x+1
check for 101.
how?
the divisibility for 101 is take last 2 digits from the right and subtract next 2 digits
i.e. (34-12)*100
= 22*100/101 = -22
so number should be of the form
9x+1 = 101y+89
hey chandrakant,
thanks for ur answer...
but can u tell me why have u multiplied 100 with (34-21)
is it because we have to do (101-1)
Thanks,
Gaurav K
Puys,isn't there a generalization to the Wilson's theorem ?
....i mean suppose we need to calculate for (p-5) ! or (p-4) ! ,don't we have an extension for such cases ?
PS: i know the normal method,if someone knows of the extension please help
ans is a.1
((38^16!)^1771)/17
=(((34+4)^16!)^1771)/17
=((4^16*15!)^1771)/17
=((16^8*15!)^1771)/17
=(((17-1)^8*15!)^1771)/17
=(((-1)^8*15!)^1771)/17
here 8 is even number and 15! is also even number because there are 3 zero at the end,
also whose last digit is zero it's whatever power is remain zero.
therefore we get even power of -1.
therefore ans is 1.
Hi Puys,
Can anyone pls explain how to solve below problem by Chinese theorem:
Whats the remainder when 123412341234......(upto 400 digits) is divided by 909.
thanks,
Gaurav K
Hey i think this is the easier way of solving this.
1234....400 times = 1234(10^(4*99)+10^(4*9
+...+10^(4*0))Now remainder of this no when divided by 909 is
rem(1234/909)*rem((10^(4*99)+..+10(4*0))/909)
= rem(325*(1+1+1+1+...100times))/909
= rem(32500/909)
=685.
Not my brain but saw it from time solutions
N=323232......................32323...if N has exactly 101 digits find the remainder when N is divied by 440
Swatysachdeva SaysN=323232......................32323...if N has exactly 101 digits find the remainder when N is divied by 440
440=40*11
& 40=8*5
N mod 8 = 3.
N mod 5 = 3.
=>N mod 40 = 3
N mod 11 = mod 11 = 53 mod 11 = 9
so remainder wld be of the form 40n+3=11m+9
440=40*11
& 40=8*5
N mod 8 = 3.
N mod 5 = 3.
=>N mod 40 = 3
N mod 11 = mod 11 = 53 mod 11 = 9
so remainder wld be of the form 40n+3=11m+9
cud u plzzz explain the last step of hw did u arrive at Nmod11==53mod9
Swatysachdeva Sayscud u plzzz explain the last step of hw did u arrive at Nmod11==53mod9
yar that is nothing but 11's divisibility rule.
sum of numbers at the odd places - sum of numbers at even.
yar that is nothing but 11's divisibility rule.
sum of numbers at the odd places - sum of numbers at even.
ohhkk .....thanks.....
Swatysachdeva SaysN=323232......................32323...if N has exactly 101 digits find the remainder when N is divied by 440
Finding answers to such questions becomes very easy if you have options.For eg. N mod 11 is 9...so the remainder (R) mod 11 must also be 9...now the only job is to find that option that gives a rem of 9 with 11. 😃
barclays_boss SaysFinding answers to such questions becomes very easy if you have options.For eg. N mod 11 is 9...so the remainder (R) mod 11 must also be 9...now the only job is to find that option that gives a rem of 9 with 11. :)
thats when ur falling short of time, coz sometimes what happens is u chk the first option, it concurs to ur 9k+xx criterion. but so does option d. so agar time ho toh fool-proof plan karo.
Ps - mere sath ho chuka h ek mock mein. 😛