k lets see,
we know that any number repeated (p-1) times is divisible by p. where p is a prime.
Thanks a lot....
Is there any condition on p?
Suppose if I take p=5 and N= 4444, Here N is not completely divisible by 5.
Thanks a lot....
Is there any condition on p?
Suppose if I take p=5 and N= 4444, Here N is not completely divisible by 5.
Thanks a lot....
Is there any condition on p?
Suppose if I take p=5 and N= 4444, Here N is not completely divisible by 5.
forgot to mention, its for p>5
What is the remainder when 12341234...upto 400 digits is divided by 909?
Approach required.
What is the remainder when 12341234...upto 400 digits is divided by 909?
Approach required.
909 = 9*101
(1+2+3+4)*100 mod 9 = 1
101 ki property hoti h that we take 2 digits at a time and substract.
34*100-12*100
2200
2200mod101=79
hence the rem will be of the form
9k+1 = 101k+79
685 fits.
What is the remainder when 12341234...upto 400 digits is divided by 909?
Approach required.
Answer : 685 ?
10^4 % 909 = 1 (% means remainder)
Therefore the big number reduces to 1234X100 % 909
or 3412 % 909
= 685
Please verify ...
k lets see,
so 888888...... till 92 digits will be completely div by 47
in the question we have 88(upto 89 digits)
so x888 mod 47 = 0
since last digit is 8 => 4*47 = 188
last two digit is taken care of
this means x8 leaves remainder 1 when divided by 47..
48 mod 47 = 1
so x = 4 ...
remainder is 4..
hope its clear.
Could you please explain the part in bold.
I got till the part of x=4 which makes the last four digits of the no. 4888
909 = 9*101
(1+2+3+4)*100 mod 9 = 1
101 ki property hoti h that we take 2 digits at a time and substract.
34*100-12*100
2200
2200mod101=79
hence the rem will be of the form
9k+1 = 101k+79
685 fits.
Answer toh correct hai but yeh funde m *100 kyu kiya??
And aise aur bhi koi funde hai kya??
Could you please explain the part in bold.
I got till the part of x=4 which makes the last four digits of the no. 4888
bhai number of digits 4888 ni.
we saw ki agar 3 extra 8 hote toh the number wld hv been prfctly div. so we took those extra 8's mod 47 = 0
abb we dnt knw what wld be the rem of 888...89 times by 47
so we assumed it to be x888 mod 47 = 0
here x is the remainder.
now 47*4 = 188
=> x8 must leave a remainder 1 with 47
aesa konsa number h?
48 mod 47 = 1 haina?
so x = 4.
hence the remainder is 4.
i hope u got it this time?
Answer toh correct hai but yeh funde m *100 kyu kiya??
And aise aur bhi koi funde hai kya??
arre 4 digits li h at a time. so 400 digits ko compensate karne ke liye *100. 😃
bhai number of digits 4888 ni.
we saw ki agar 3 extra 8 hote toh the number wld hv been prfctly div. so we took those extra 8's mod 47 = 0
abb we dnt knw what wld be the rem of 888...89 times by 47
so we assumed it to be x888 mod 47 = 0
here x is the remainder.
now 47*4 = 188
=> x8 must leave a remainder 1 with 47
aesa konsa number h?
48 mod 47 = 1 haina?
so x = 4.
hence the remainder is 4.
i hope u got it this time?
Thanks bhai...
Just another small koschin... supposedly we were asked to find the remainder of 888...(87) times when divided by 47?
Does the approach change for finding the remainder (i.e digit x in x88888 ).. I tried carrying forward the funda of 4888 being completely divisible by 47 thus to find x such that x88 leaves a remainder of 4... but am unable to get a value for x
Thanks bhai...
Just another small koschin... supposedly we were asked to find the remainder of 888...(87) times when divided by 47?
Does the approach change for finding the remainder (i.e digit x in x88888 ).. I tried carrying forward the funda of 4888 being completely divisible by 47 thus to find x such that x88 leaves a remainder of 4... but am unable to get a value for x
Bhai i think here x can be any number less than 47 its not necessary for it to be a single digit. so if you check for '33'88 mod 47 = 4 and hence remainder is 33
Enceladus bhai please check...
Thanks bhai...
Just another small koschin... supposedly we were asked to find the remainder of 888...(87) times when divided by 47?
Does the approach change for finding the remainder (i.e digit x in x88888 ).. I tried carrying forward the funda of 4888 being completely divisible by 47 thus to find x such that x88 leaves a remainder of 4... but am unable to get a value for x
bhai its not 4888 completely being div by 47. we just had to look for a 2 digit number (x8 in our case) which wld leave a remainder 1 with 47. isliye answer 4 aya humara.
Bhai i think here x can be any number less than 47 its not necessary for it to be a single digit. so if you check for '33'88 mod 47 = 4 and hence remainder is 33
Enceladus bhai please check...
han bhai x can be any number depending upon our condition. 😃
Find greatest two digit prime number, which will divide 200C100 perfectly....
Paritosh1003 SaysFind greatest two digit prime number, which will divide 200C100 perfectly....
It should be 61.
200C100 = 200!/(100!*100!)
= 101*102*103*...*200/(1*2*3*..*100)
Now, we need to find out largest prime number less than 100 such that it has at least 2 multiple of it in numerator i.e. 2p and 3p.
So, 3p p can be at most 61.
Here, 61 will have 2 multiples in numerator, 122 and 183.
61 in denomerator will cancel out one of them. Remaining one can be divided by 61.
This question is already discussed before, but I didn't understand.
What is the remainder when 5555......93times divided by 98?
This question is already discussed before, but I didn't understand.
What is the remainder when 5555......93times divided by 98?
N = 555.... 93 times
98 = 2*49
Nmod2 = 1
N = 5/9 ( 10^93 - 1)
E = 42
49 is coprime to 10. So 10^42Mod49 = 1,
Since 49 is coprime to 9 as well, we have
Nmod49 = {5/9 (10^93 - 1)}Mod49 = {5/9(10^9 -1 )}Mod49
5/9 (10*2^4 - 1)Mod49 = 5/9(159)mod49 = 5/9 * 306Mod49 = 5*34Mod49 = 170Mod49 = 23
So, Remainder is of the form , 49x + 23 = 2y + 1
@x = 0, y = 11 we have Rem = 23
Part highlighted in Blue --> Basically 159Mod49 = 306Mod49 = 12. We change it to 306, so that it is divisible by the denominator '9'.
look dude,you can by two methods..
1st
let no of days=x
work done per day=y
now we have two equations
x*y=1800 and
x/3*(y-20)+(2x-3/3)*(y+20)=1800
solving we get
20x=3y+60
so y=20x-60/3
putting y in x*y=1800
we get x=18
so y=100
2nd method.
Just go by options.
Suppose take answer as 100
then cross check.
No of day=1800/100
=18
so as per questn
in 6 days
6*80=480tons work is done.
So now in 11 days(wrk is cmpltd bfr 1 day)
work done should be
1800-480=1320.
Check for 11*120
it comes as 1320.
So 100 is correct
is it 20 tons overall for (x/3) days or 20tons per day less? ? how did we interpret that line?
imsheela Saysis it 20 tons overall for (x/3) days or 20tons per day less? ? how did we interpret that line?
In the next line it is given that ''To make up for this, the team overachieved for the rest of the days by 20 tons'' ..so from this you can interpret that 20 tons per day was less 😃
Im having little bit of trouble applying Euler's Theorem in certain questions. I would like a simplified explanation of how to use it. eg R etc 
Im having little bit of trouble applying Euler's Theorem in certain questions. I would like a simplified explanation of how to use it. eg R[7^103/100] etc
I will try to explain euler's theorem:
EULERS'S THEOREM :
given: to find remainder when M^a is divided by N
NOTE: euler's theorem applicable only if M and N are co-prime to each other(or have no factors in common in oder words)
the theorem says that the remainder of M^K wen divided by N is always 1 where k= N (1-(1/x) (1-(1/y) (1-(1/z))...here x,y,z are the primes u obtain wen u write N in standard form
to simplify things i will take ur question: 7^103 wen divided by 100.
now 1st step is to check wedr M and N are co-prime to each other. 7 and 100 ar co-prime. So, i can safely apply euler's theorem
2nd step :find "k"..
100=(2^2) x (5^2)
so k = 100 (1-(1/2) (1-(1/5)= 40
thus EULER'S THEOREM says that remainder of 7^40 wen divided by 100 is 1, or R((7^40)/100) =1
now question is R((7^103)/100)=R[ (7^40)(7^40)(7^23) /100 ]=R [7^23/100] = 43 (since R(7^40) (7^40) / 100) = 1 X 1=1
hope the explanation helps
#dreamer-on-the-rocks#