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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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