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 101 is a prime. 33 solutions Vote up solutions you admire Correct answer: 19 Krutarth Patel 22, India Solution writing guide: Level 4 Formatting guide PREVIEW Write a solution. Insert an image 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 Reply Subscribe to comments Calvin Lin STAFF 30, USA Jul 16, 2013 Nice solution. Note that you didn't need to use Wilson's Theorem at all, since you already pointed out that the cancellation will occur. This is very similar to the proof of Lucas Theorem. Here's a vote boost from me :) 2 0 Reply Sambit Senapati 19, India Jul 17, 2013 In response to Calvin Lin: Yes, you are right. :) Next problem → 1 Home Stats Community Topics Search
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7/16/2015 (1) Brilliant
https://brilliant.org/ 1/12
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
Vote up solutions you admire
Correct answer: 19
Krutarth Patel
22, India
Solution writing guide: Level 4
Formatting guide
PREVIEW
Write a solution.
Insert an image
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.
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:
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:
(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 .
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)
Reply
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.
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 .
Reply
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 .
Reply
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)
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
Reply
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)
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.
Reply
Harsa Mitra
21, India Upvote (0) Jul 16, 2013
Using Lucas Theorem (http://en.wikipedia.org/wiki/Lucas'_theorem)