(mp + n)cp solution
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Problems of the Day
← What's that remainder? · Level 4
(175 points)
What is the remainder, when is divided by 101?
Details and assumptions
You may use the fact that 101is a prime.
33 solutions
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Correct answer: 19
Krutarth Patel
22, India
Solution writing guide: Level 4
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Sambit Senapati
19, India Upvote (45) Jul 15, 2013
My solution uses Wilson's theorem. Here it is:
Let (where )
Multiply both sides of the equation by 100!.
By wilson's theorem
2013,2012,. . ., 1920 give remainders 94,93, . . ., 1 respectively when divided by 101.
1913,1914, . . .,1918 give remainders 95,96, . . ., 100 respectively on division by 101.
So,
Therefore
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Calvin Lin STAFF
30, USAJul 16, 2013
Nice solution. Note that you didn't need to use Wilson's Theorem at all, since youalready pointed out that the cancellation will occur. This is very similar to the proofof Lucas Theorem.
Here's a vote boost from me :)
2 0 Reply
Sambit Senapati
19, IndiaJul 17, 2013
In response to Calvin Lin: Yes, you are right. :)
Next problem →
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1 0 Reply
Ruskin Bond
19, United KingdomJul 16, 2013
An excellent method. Like it more than Lucas.
2 0 Reply
Jean Lee
19, South KoreaJul 24, 2013
Whoa. NICE.
1 0 Reply
Abul Ahmed
15, BangladeshJul 23, 2013
1 0 Reply
Yunhao King
17, SingaporeJul 18, 2013
No Lucas!Nice!
1 0 Reply
Sri Kanth
28, IndiaJul 15, 2013
I think Lucas theorem is an overkill. This solution is really nice :)
1 0 Reply
Hero P.
39, USA Upvote (18) Jul 15, 2013
An elementary proof without resorting to Lucas' Theorem follows. First, we claim forall primes , positive integers , and integers ,
For , the claim is trivially true. For ,
But since the LHS is an integer, is prime, and cannot divide under
the given conditions, it follows that it must divide . So the first term on
the RHS is divisible by . Hence
and induction on proves our claim. Next, we claim that for positive integers ,
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For , the claim is again trivial: . For , we write
Thus by the first claim with ,
By induction on and recalling that implies for prime , the result immediately follows. Therefore, we have proved
and with the choice , we find leaves a remainder
of when divided by .
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Muhammad Al Kahfi
18, Indonesia Upvote (18) Jul 15, 2013
First of all, let we see the Lucas theorem :http://ecademy.agnesscott.edu/~lriddle/ifs/siertri/LucasProof.htm
Now, since
Then, by Lucas theorem above, we obtain :
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Rindell Mabunga
17, PhilippinesJul 18, 2013
wow amazing i did not know that there was an existing theorem like that
2 0 Reply
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Vishwa Iyer
16, USAJul 15, 2013
you should change your and into and
respectively.
(last modified Jul 15, 2013 by a moderator)
1 0 Reply
Jimmy Kariznov
25, USAJul 15, 2013
In response to Vishwa Iyer: In my opinion, \cdot is better for denotingmultiplication than . or *, but that's just me.
3 0 Reply
Sotiri Komissopoulos
19, USA Upvote (10) Jul 15, 2013
. We see that .
Considering the rest of the numerator (without the ) , we have each ofthe integers between and , inclusive, present. That is,
, since for any prime , by Wilson'sTheorem (for more information, see http://en.wikipedia.org/wiki/Wilson's_theorem).Similarly, looking at the rest of the denominator (without the ), we have
.
With these in mind, we can solve our problem:
. Therefore, the
remainder when is divided by is .
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Kiriti Mukherjee
17, Bangladesh Upvote (7) Jul 15, 2013
According to lucas theorem it can be solved.. since applying lucas theorem-
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Jimmy Kariznov
25, USA Upvote (6) Jul 15, 2013
Note that and . Then, since is prime,by Lucas' Theorem, we have:
.
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George Williams Upvote (5) Jul 16, 2013
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George Williams
21, USA Upvote (5) Jul 16, 2013
This is a nice solution, the source of the corresponding exercise is Apostol's. For all ,we have that:
Note that this is a special case of Lucas theorem, which has already been describedhere.
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Alyosha Latyntsev
19, United KingdomJul 22, 2013
Do you have the name of this theorem?
1 0 Reply
Yunhao King
17, SingaporeJul 18, 2013
I can prove this without Lucas therorem. Inside p consecutive integers k,k-1,...k--p+1,there must have an integer can bedivided by p,we let this integer be k-i. 0 =<i<=p-1;then we have [k/p]=[k-i+i/p]=k-i/p+[i/p]=k-i/p
Let Q=(kk-1k-2...k-p+1)/k-i; Then we have Q≡(p-1)!(mod p);
And Q[k/p]=Q(k-i)/p=(p-1)!{k \choose p}; (p-1)![k/p]≡Q[k/p]≡(p-1)!{k \choose p}(mod p); As p is a prime number;(p-1)! can not be divided by p;as a result {k \choose p}=k/p
(last modified Jul 18, 2013 )
1 0 Reply
Yunhao King
17, SingaporeJul 18, 2013
Wow!!!!I like this one!
1 0 Reply
Mayank Kaushik
21, IndiaJul 16, 2013
I did the same as you did , But I didn't know that this result is the special case ofLucas Theorem (which i never heard)
1 0 Reply
Oscar Harmon
18, USA Upvote (4) Jul 15, 2013
We can see that . Now, we could rearrange this to be
, but these are not precisely
equivalent since . So, we can factor this out initially and proceed:
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Calvin Lin STAFF
30, USAJul 16, 2013
Indeed, you have to be careful with what means. Thanks for
pointing that out!
1 0 Reply
Daniel Chiu
16, USA Upvote (3) Jul 15, 2013
We must find this modulo 11.
Notice the top is equal to modulo 101. Now, the cancel, and the
answer is
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Xuming Liang
17, USA Upvote (2) Jul 19, 2013
Here's a solution using the simple lemma:
when are relatively prime.
Proof: for some integer , , since are relatively prime, thus it's clear that
. Q.E.D
Now consider the number , which is equal to
, thus we have
. Since is prime, thus and are
relatively prime. By the lemma above we have
, which
means that the remainder is .
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Atonu Mukherjee
20, Bangladesh Upvote (2) Jul 15, 2013
According to lucas theorem it can be solved.. since
applying lucas theorem-
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Ang Yan Sheng
30, Singapore Upvote (1) Jul 21, 2013
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30, Singapore
Note that
Hence
(Note: all the divisions are valid (mod 101) because the denominators of every fraction,except the first, are coprime to 101.)
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Abhishek Pushp
16, India Upvote (1) Jul 20, 2013
it can be solved by wid lucas theorem now, 2013=19.101^1+94. 101=1.101^1+0 byLucas theorem : m1=19 m0=94 n1=1 n2=0 C(2013 101)≡C(19 1).C(94 0) ≡19(mod101) SO 19 IS ANSWER
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Hesto Plowkeeper
15, USA Upvote (1) Jul 17, 2013
Without resorting to fancy combinatoric theorems, we use generating function(1+x)^2013, and consider the coefficient of term x^101, which is 2013C101. Undermod 101, (1+x)^101=1+x^101, for all the coefficients in the middle are divisible by101. Hence (1+x)^2013 = (1+x^101)^19 (1+x)^94, of which the coefficient of x^101is 19 by binomial expansion. Namely, the answer is 19.
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Jorge Fernández
19, Mexico Upvote (0) Jul 12
, top and bottom is a complete residue system, the multiples of are
and . They make when dividing. The rest of the complete residuesystem cancels out .
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Patrick Corn
39, USA Upvote (0) May 20, 2014
We have after canceling a factor
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We have after canceling a factor
of from top and bottom.
The numbers left in thenumerator run through all of the nonzero congruence classes mod exactly once.So, modulo , the numerator is congruent to . Since is prime, isinvertible mod , so it makes sense to rewrite:
mod .
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Rahul Nahata
19, India Upvote (0) May 20, 2014
According to Lucas' theorem since and Therefore
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Ayush Saini
21, India Upvote (0) May 20, 2014
My solution uses Wilson's theorem. Here it is:
Let (2013101)=2013×2012×…×1913(101)!≡c(mod101) (where 0≤c<101)
Multiply both sides of the equation by 100!.
By wilson's theorem 100!≡−1(mod101) ⇒2013×2012…×(1919101)…×1913≡c×100!≡−c(mod101) 2013,2012,. . ., 1920 give remainders 94,93, . . ., 1 respectively whendivided by 101.
1913,1914, . . .,1918 give remainders 95,96, . . ., 100 respectively on division by 101.
So, (2013101)≡94!×19×100×99…×95≡19×100!≡−c(mod101) ⇒c=19 Therefore(2013101)≡19(mod101)
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Bờ La Bốc Khói
18, Vietnam Upvote (0) May 20, 2014
Since 2013=19.101+94 then according to the lucas theorem we have ${2013 \choose101} \equiv {19 \choose 1}{94 \choose 0} \equiv 19 \pmod{101}$ Ans: 19.
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Douglas Zare
38, USA Upvote (0) May 20, 2014
The first term simplifies: . The other terms cancel in the arithmetic of the
integers mod 101, since each numerator is congruent to the denominator mod 101.So, the product is 19 mod 101.
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So, the product is 19 mod 101.
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Ashwath Thirumalai
17, Uganda Upvote (0) May 20, 2014
By Lucas' Theorem, since 101 is prime, this binomial coefficient is equal to a bunch ofstuff choose 0 (which is just 1) times 19 choose 1 (modulo 101). This is clearly 19 mod101.
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Thomas Baxter
21, Canada Upvote (0) Jul 21, 2013
The question is equivalent to, "What is equivalent to modulo , as a
number from to inclusive?"
View this combination as the product .
Note that the first fraction on the right-hand side is equivalent to modulo ,because the top and bottom each include a number equivalent to modulo foreach integer , so their modular products are equal and non-zero.
Then, modulo , the combination is equivalent to .
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Jason Martin
23, USA Upvote (0) Jul 21, 2013
We know . In mod 101, we can treat division by 100, 99,
98, etc as multiplication by their multiplicative inverses. Thus, we have
. Each value from 1913 to 2013 is a value
mod 101. The only value that is divisible by 101 is 1919. Thus, everything cancels
until we're left with .
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Christopher Boo
18, Malaysia Upvote (0) Jul 20, 2013
This is a typical problem to be solved by Lucas Theorem. (The proof and explanationof Lucas Theorem is too complicated, I will only write the way to tackle this problem,for more information you can Google it)
Next,
So, the remainder of when divided by is .
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Tran Trung Nguyen
18, Vietnam Upvote (0) Jul 20, 2013
2013=19.101^1+94. 101=1.101^1+0 by Lucas theorem : m1=19 m0=94 n1=1 n2=0C(2013 101)≡C(19 1).C(94 0) ≡19(mod 101) SO 19 IS ANSWER
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Evan Chien
16, USA Upvote (0) Jul 19, 2013
2013=19.101^1+94
101=1.101^1+0
by Lucas theorem:
m1=19
m0=94
n1=1
n2=0
C(2013 101)≡C(19 1).C(94 0)
≡19(mod 101) So 19 is the answer
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Utsav Singhal
16, India Upvote (0) Jul 18, 2013
It means.....2013 c 101 = 2013! ____ 101! (2013-101)! solve this and divide it by 101which will give the remainder 19
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Abhishek Srivastava
18, India Upvote (0) Jul 18, 2013
2013=19.101^1+94. 101=1.101^1+0 by Lucas theorem : m1=19 m0=94 n1=1 n2=0C(2013 101)≡C(19 1).C(94 0) ≡19(mod 101) SO 19 IS ANSWER
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Ajay Kumar
20, India Upvote (0) Jul 18, 2013
by seeing the question we will aply Lucas theorem now 2013=19.1011+94.101=1.1011+0 Then, by Lucas theorem above, we obtain : m1=19 m0=94 n1=1 n2=0(2013101)≡(191).(940)≡19.1≡19(mod101)
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Tan Likai
17, Singapore Upvote (0) Jul 17, 2013
Since . Note that . Then note that the
numbers from 1913 to 2013 exculding 1919 are 1 to 100 modulo 101. By Wilson's
Theorem,
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Abhishek Kumar
17, India Upvote (0) Jul 16, 2013
According to lucas theorem it can be solved.. since applying lucas theorem-
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Debjit Mandal
20, India Upvote (0) Jul 16, 2013
I am going to use the result that, if p is a prime and m and n are positive integerssatisfying m=ap+r , n=bp+s where 0\leq r,s<p, then {m \choose n}≡{a \choose b}{r\choose s}\pmod{p}. Here, m=2013= 101 \times 19 + 94, and n=101 \times 1 + 0. So,{2013 \choose 101}≡{19 \choose 1}{94 \choose 0}≡19 \times 1≡19\pmod{101}. So,the answer is 19.
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Harsa Mitra
21, India Upvote (0) Jul 16, 2013
Using Lucas Theorem (http://en.wikipedia.org/wiki/Lucas'_theorem)
2013=19.101+94. 101=1.101+0
Using Lucas theorem:-
we get, 19 (mod101)
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