Brilliant Junior Number Theory Contest (Season 1) ENDS

Hello every Brilliant Mind! This is the first season of Brilliant Sub Junior Number Theory Contest. This contest is for beginners or intermediate ones who want to sharpen their skills of problem-solving in overall Number Theory. .

The aim of this contest is to improve the skill of in the computation in all sorts of problem (of basic level) in Number Theory like Proof problems , Types of Congruences , Theorems related to Modular Arith. etc. by learning from each other and of course to have fun !

Eligibility

People should fulfill either of the 2 following:

  1. 16 years or below

  2. Minimum level preference : Level 3 in Number Theory .


Eligible people here may participate in this contest.

The rules are as follows : \text{The rules are as follows : } :

(There is a slight change in the rules of the contest : Now there can be more(2-3) unsolved problems(they are called 'Extra Participation Problem(EPP)) ! , so many people can participate in contest at once ! :) One rule to be noted about these EPP's is that , those problems are to be posted until regular problem is unsolved.For each new unsolved problem , there will be new set of EPP's ! Happy Solving !)

  1. I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  2. You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  3. Only make substantial comment that will contribute to the discussion.

  4. Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  5. If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  6. It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant..

  7. Please post your solution and your proposed problem in a single new thread.


Format your post is as follows :\text{Format your post is as follows :}

SOLUTION OF PROBLEM xxx (number of problem) :

[Post your solution here]

PROBLEM xxx (number of problem) :

[Post your problem here]]]

[Name of the solver]


The comments will be easiest to follow if you sort by "Newest":


Topics allowed for the contest are:\text{Topics allowed for the contest are:}

Proofs related to congruences , Number Theoretic inequalities , Eulers Theorem , Modular Arithmetic , Number Theoretic Functions , Order of an Integer modulo nn .


Things to keep in mind:\text{Things to keep in mind:}

You may include other pure NT topics.
Try to avoid toughest problems , such as problems related to Dirichlet's Convolution etc. as this is Junior Contest , the second season will be totally reserved for hard and toughest problems which community will enjoy .
Targeted participants for this contest are beginners .

You can discuss about the problems on\text{You can discuss about the problems on} Discussion Channel for BJNTC

Note by A Former Brilliant Member
3 years, 7 months ago

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1 vote

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d1,d2,.....d2nd_{1},d_{2},.....d_{2n} can be from 1,2,3,4,...,91,2,3,4,...,9 only

j=1nd2j1d2j\sum_{j=1}^{n}d_{2j-1}d_{2j} is even If

(1)(1) all the nn terms are even

(2)(2) the number of odd terms in the sum is even

d2j1d2jd_{2j-1}d_{2j} is even if

(1)(1) both d2j1d2jd_{2j-1}d_{2j} are even . There are 16 choices for this

(2)(2) one of them is odd the other one is even ; there are 2(5)(4)=402(5)(4)=40 choices for this d2j1d_{2j-1} could be odd and d2jd_{2j} even or vice versa ).

Note d2j1d2jd_{2j-1}d_{2j} is odd for 2525 choices .

For the sum to be even , an even number of such products must be odd . Number of ways in which kk of the nn terms are odd and the rest are even is (nk).25k(56)nk\binom{n}{k}.25^{k}(56)^{n-k}.

Total sum is even if ksk's are even .

Total number of ways

Pn=k=0n2(n2k)252k56n2kP_{n}=\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}\binom{n}{2k}25^{2k}56^{n-2k}

Here.\lfloor.\rfloor represents floor function.

Let Qn=k=0n2(n2k+1)252k+156n2k1Q_{n}= \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}\binom{n}{2k+1}25^{2k+1}56^{n-2k-1}

Pn+Qn=(56+25)nP_{n}+Q_{n}=(56+25)^{n}

PnQn=(5625)nP_{n}-Q_{n}=(56-25)^{n}

Hence Pn=81n+31n2\boxed{P_{n}=\frac{81^{n}+31^{n}}{2}}

Shivam Jadhav - 3 years, 7 months ago

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Nice solution.

Sharky Kesa - 3 years, 7 months ago

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Awesome ! nice solution ....

A Former Brilliant Member - 3 years, 7 months ago

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Post next regular problem ...

A Former Brilliant Member - 3 years, 7 months ago

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Problem 2

Prove that all numbers in the form 224n+1+72^{2^{4n+1}}+7 and 226n+2+132^{2^{6n+2}}+13 for nNn\in \mathbb{N} are composite.

This problem has been solved by Julian Poon.

Akshat Sharda - 3 years, 7 months ago

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Solution to Problem 2

Since the 2 questions are so similar, I'll just prove for one case.

From Euler's Theorem,

210n1 (mod11)2^{10n}\equiv 1\ \left(\operatorname{mod}11\right)

So,

224n+1 2[24n+1 (mod 10)]mod 11 2^{2^{4n+1}}\ \equiv 2^{\left[2^{4n+1}\ \left(\operatorname{mod}\ 10\right)\right]}\operatorname{mod}\ 11\

Using Euler's theorem again,

24n+12 (mod 10)2^{4n+1} \equiv 2\ (\operatorname{mod}\ 10)

Hence

2[24n+1 (mod 10)]=4 (mod 11)2^{\left[2^{4n+1}\ \left(\operatorname{mod}\ 10\right)\right]}=4\ (\operatorname{mod}\ 11)

The result follows, that 224n+1+72^{2^{4n+1}}+7 is divisible by 11.

Similarly, you can prove that 226n+2+132^{2^{6n+2}}+13 is divisible by 2929.

Julian Poon - 3 years, 7 months ago

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@Akshat Sharda Can it be solved this way that all primes are of the form 6k + 1 or 6k - 1. Its easy to show that they are not of the either forms and hence they are composite???

Ankit Kumar Jain - 3 years, 7 months ago

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6(8)+1=49 and 6(6)-1=35

49 and 35 are composite.

Akshat Sharda - 3 years, 7 months ago

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@Akshat Sharda @Akshat Sharda what I meant is that all primes are of the form 6k +- 1 and not that all 6k+-1 are primes

Ankit Kumar Jain - 3 years, 7 months ago

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@Ankit Kumar Jain Not all primes.... What about 2 and 3?

Akshat Sharda - 3 years, 7 months ago

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@Akshat Sharda @Akshat Sharda Leaving 2 and 3...I forgot to mention that....Btw Akshat you are in 9 or 10??

Ankit Kumar Jain - 3 years, 7 months ago

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@Ankit Kumar Jain I'm in 10th grade.

Akshat Sharda - 3 years, 7 months ago

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No, because all numbers in the form above are -1, 1 mod 6

Julian Poon - 3 years, 7 months ago

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@Julian Poon @Julian Poon O sorry I didn't mark that the numbers above are of the form 6k-1

Ankit Kumar Jain - 3 years, 7 months ago

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Problem 6

Find all positive integers (with proof) x,y,zx,y,z such that 8x+15y=17z8^{x}+15^{y}=17^{z}

This problem has been solved by Sharky Kesa.

Shivam Jadhav - 3 years, 7 months ago

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Here is my extremely long method to solving this question (First method I found) :P

8x+15y=17z1+117z(mod7)17z2(mod7)z 2(mod6)\begin{aligned} 8^x + 15^y &= 17^z\\ \Rightarrow 1+1 &\equiv 17^z \pmod{7}\\ \Rightarrow 17^z &\equiv 2 \pmod{7}\\ \Rightarrow z\ &\equiv 2 \pmod{6} \end{aligned}

Thus, zz is even. Also, we have

8x+15y=17z8x=17z15y0=1x(1)y(mod8)y=0(mod2)\begin{aligned} 8^x + 15^y &= 17^z\\ 8^x &= 17^z - 15^y\\ 0 &= 1^x - (-1)^y \pmod{8}\\ y &= 0 \pmod{2} \end{aligned}

Thus, yy is even. Also, we have (from looking at the equation in (mod10)\pmod{10}), 84a+15b=174c8^{4a}+15^b = 17^{4c} and 84a+2+15b=174c+28^{4a+2} + 15^b = 17^{4c+2}. Thus, xx is even. Let x=2x1x=2x_1, y=2y1y=2y_1 and z=2z1z=2z_1. Substituting back in, we get

26x1=172z1152y126x1=(17z115y1)(17z1+15y1)\begin{aligned} 2^{6x_1} &= 17^{2z_1} - 15^{2y_1}\\ 2^{6x_1} &= (17^{z_1} - 15^{y_1})(17^{z_1} + 15^{y_1}) \end{aligned}

Because of FTA, we can let 17z115y1=2a17^{z_1} - 15^{y_1}=2^a and 17z1+15y1=2b17^{z_1} + 15^{y_1} = 2^b, with a+b=6x1a + b = 6x_1 and a<ba < b. We now have

17z115y1=2a17z1+15y1=2b17y1=2b1+2a1\begin{aligned} 17^{z_1} - 15^{y_1} &= 2^a\\ 17^{z_1} + 15^{y_1} &= 2^b\\ \Rightarrow 17^{y_1} &= 2^{b-1} + 2^{a-1} \end{aligned}

From this, since the LHS is odd, the RHS must be odd, so a=1a = 1 and b=6x11b = 6x_1 - 1. Substituting this, we get

17z115y1=217z1+15y1=26x11(1)z1+01(mod3)z11(mod2)\begin{aligned} 17^{z_1} - 15^{y_1} &= 2\\ 17^{z_1} + 15^{y_1} &= 2^{6x_1-1}\\ \Rightarrow (-1)^{z_1} + 0 &\equiv -1 \pmod{3}\\ \Rightarrow z_1 &\equiv 1 \pmod{2} \end{aligned}

17z115y1=217z1+15y1=26x111z1(1)y10(mod8)y11(mod2)\begin{aligned} 17^{z_1} - 15^{y_1} &= 2\\ 17^{z_1} + 15^{y_1} &= 2^{6x_1-1}\\ \Rightarrow 1^{z_1} - (-1)^{y_1} &\equiv 0 \pmod{8}\\ \Rightarrow y_1 &\equiv 1 \pmod{2} \end{aligned}

Thus, y1y_1 and z1z_1 are odd, which implies that x1x_1 is odd from 84a+2+15b=174c+28^{4a+2} + 15^b = 17^{4c+2}.

Going back to the original equation and substituting x=4x2+2x=4x_2+2, y=4y2+2y=4y_2+2 and z=4z2+2z=4z_2+2, we get

84x2+2+154y2+2=174z2+2154y2+2=(172z2+182x2+1)(172z2+1+82x2+1)\begin{aligned} 8^{4x_2+2} + 15^{4y_2+2} = 17^{4z_2+2}\\ 15^{4y_2+2} = (17^{2z_2+1} - 8^{2x_2+1})(17^{2z_2+1} + 8^{2x_2+1}) \end{aligned}

Once again, by FTA, let 172z2+182x2+1=3a5b17^{2z_2+1} - 8^{2x_2+1} = 3^a 5^b and 172z2+1+82x2+1=3b5a17^{2z_2+1} + 8^{2x_2+1}=3^b 5^a, where a+b=4y2+2a+b=4y_2+2.

172z2+182x2+1=3a5b172z2+1+82x2+1=3b5a(1)2z2+1+(1)2x2+10(mod3)(1)10(mod3)\begin{aligned} 17^{2z_2+1} - 8^{2x_2+1}&= 3^a 5^b\\ 17^{2z_2+1} + 8^{2x_2+1}&=3^b 5^a\\ \Rightarrow (-1)^{2z_2+1} + (-1)^{2x_2+1} &\equiv 0 \pmod{3}\\ \Rightarrow (-1)-1 &\equiv 0 \pmod{3} \end{aligned}

Since this isn't true, we have that b=0b=0 so a=4y2+2a=4y_2+2. We now have

172z2+182x2+1=92y2+1172z2+1+82x2+1=252y2+1\begin{aligned} 17^{2z_2+1} - 8^{2x_2+1}&= 9^{2y_2+1}\\ 17^{2z_2+1} + 8^{2x_2+1}&= 25^{2y_2+1} \end{aligned}

For x2>0x_2 > 0, we have 82x2+10(mod64)8^{2x_2+1} \equiv 0 \pmod{64}. Also, 172z2+117,49(mod64)17^{2z_2+1} \equiv 17, 49 \pmod{64} and 252y2+125,9,57,4125^{2y_2+1} \equiv 25,9,57,41. Since both of these terms don't share any residues in (mod64)\pmod{64}, x2=0x_2 = 0. Substituting this back in to the original equaition, we get

82+15y=17z17z1+15y1=3217z115y1=217z1=1715y1=15z1=1y1=1\begin{aligned} 8^2 + 15^y &= 17^z\\ \Rightarrow 17^{z_1} + 15^{y_1} &= 32\\ 17^{z_1}-15^{y_1} &= 2\\ \Rightarrow 17^{z_1} = 17\\ 15^{y_1} = 15\\ \Rightarrow z_1&=1\\ y_1&=1 \end{aligned}

Thus, x=2x=2, y=2y=2, z=2z=2 is the only solution.

Sharky Kesa - 3 years, 7 months ago

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I think that if we work with modulo 32 or 64 the solution would be shorter. I haven't solved the problem yet but I think that there are some interesting results with it.

A Former Brilliant Member - 3 years, 7 months ago

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I have a similar (and equally long) solution and was hoping for a shorter one but... oh well. By the way, I think you have a typo where 1z1(1)y10mod31^{z_1}-(-1)^{y_1}\equiv 0 \operatorname{mod}3 should be 1z1(1)y10mod41^{z_1}-(-1)^{y_1}\equiv 0 \operatorname{mod}4

Julian Poon - 3 years, 7 months ago

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i have a very short solution to this question using Beals conjecture only if am permited to post it

Benjamin ononogbu - 3 years, 7 months ago

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@Benjamin Ononogbu Beals conjecture is still technically unproven, otherwise I'd have used that as my solution.

Sharky Kesa - 3 years, 7 months ago

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Are integers x,y,z are different?

Akshat Sharda - 3 years, 7 months ago

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If they are not different , then first comes pythagoras XD

A Former Brilliant Member - 3 years, 7 months ago

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Problem 1

n=1Mp.(ϕ(1+pn1))n=1Mpn+1pn \sum_{n=1}^{M}p. (\phi(1+p^{n-1})) \leq \sum_{n=1}^{M} p^{n+1}-p^n

Disprove (without giving examples , give a proof), for natural numbers MM and pp.

This problem has been solved by Akshat Sharda

A Former Brilliant Member - 3 years, 7 months ago

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Solution to Problem 1

Let us consider,

n=1Mp.ϕ(1+pn1)n=1Mpn+1pnpn=1Mϕ(1+pn1)(p1)n=1Mpnpn=1Mϕ(1+pn1)(p1)ppm1p1n=1Mϕ(1+pn1)pm1\displaystyle \sum_{n=1}^{M}p. \phi(1+p^{n-1}) \leq \displaystyle \sum_{n=1}^{M} p^{n+1}-p^n \\ p \displaystyle \sum_{n=1}^{M} \phi(1+p^{n-1}) \leq (p-1) \displaystyle \sum_{n=1}^{M} p^n \\ p \displaystyle \sum_{n=1}^{M} \phi(1+p^{n-1}) \leq (p-1)\cdot p \cdot \frac{p^m-1}{p-1} \\ \displaystyle \sum_{n=1}^{M} \phi(1+p^{n-1}) \leq p^m-1

Now, as pNp \in \mathbb{N}, for p=1p=1,

n=1Mϕ(1+pn1)=n=1Mϕ(1+1n1)=M\displaystyle \sum_{n=1}^{M} \phi(1+p^{n-1})=\displaystyle \sum_{n=1}^{M} \phi(1+1^{n-1})=M

And,

pM1=1M1=0p^M-1=1^M-1=0

As MM is a natural number it should be obviously more than 00, but inequality is giving us M0M\leq 0. Thus , it does not hold true.

Akshat Sharda - 3 years, 7 months ago

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Extra Participation Problem - Set 1 - #3

Let m,nNm,n\in \mathbb {N} such that

lcm(m,n)+gcd(m,n)=m+n\text{lcm}(m,n)+\gcd(m,n)=m+n

Prove that one of the 2 numbers is divisible by the other.

This problem has been solved by Harsh Shrivastava.

Akshat Sharda - 3 years, 7 months ago

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Let gcd(m,n)=k

Then LCM(m,n) = (mn)/k

(mn/k) + k =m+n

On rearranging, k2(a+b)k+ab=0k^2 -(a+b)k +ab=0

Thus by quadratic formula, k = a or b

Hence proved.

Harsh Shrivastava - 3 years, 7 months ago

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Extra Participation Problem -Set 2: #2

Are n5+5n^5+5 and (n+1)5+5(n+1)^5+5 coprime for all positive integers nn? If not, find all values of nn for which this is not true.

Sharky Kesa - 3 years, 7 months ago

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Here is my long, sort of bashy solution:

Suppose that dd is a common factor of n5+5n^5+5 and (n+1)5+5(n+1)^5+5. Thus, we have the following:

d(n+1)5+5(n5+5)=5n4+10n3+10n2+5n+1d(5(n5+5)(n2)(5n4+10n3+10n2+5n+1)=10n3+15n2+9n+27d4(5n4+10n3+10n2+5n+1)(2n1)(10n3+15n2+9n+27)=7n243n23d98(10n3+15n2+9n+27)(140n+1070)(7n243n23)=50112n+27256d1569507840(7n243n23)(219240n1466005)(50112n+27256)=3858751960d3858751960\begin{aligned} d & \mid (n+1)^5+5 - (n^5+5) = 5n^4 + 10n^3 + 10n^2 + 5n + 1\\ \Rightarrow d & \mid (5(n^5 + 5) - (n - 2)(5n^4 + 10n^3 + 10n^2 + 5n + 1) = 10n^3 + 15n^2 + 9n + 27\\ \Rightarrow d & \mid 4(5n^4 + 10n^3 + 10n^2 + 5n + 1) - (2n-1)(10n^3 + 15n^2 + 9n + 27) = 7n^2-43n-23\\ \Rightarrow d & \mid 98(10n^3 + 15n^2 + 9n + 27) - (140n + 1070)(7n^2-43n-23) = 50112n + 27256\\ \Rightarrow d & \mid 1569507840(7n^2 - 43n - 23) - (219240n - 1466005)(50112n + 27256) = 3858751960\\ \Rightarrow d & \mid 3858751960 \end{aligned}

The prime factorisation of 3858751960 is 235719687512^3 \cdot 5 \cdot 7 \cdot 1968751 (you can check the last term is a prime though any means of primality checking).

Since n5+5n^5+5 and (n+1)5+5(n+1)^5+5 are alternatively odd and even, 22 cannot be their common factor. Also, by FLT, since n5n(mod5)n^5 \equiv n \pmod{5}, dd cannot be equal to 5. By inspection, it can be seen that n5n^5 is not congruent to (n+1)5+5(n+1)^5+5 in (mod7)\pmod{7}. Thus, d=1d=1 or d=1968751d=1968751.

We will now find when the gcd(n5+5,(n+1)5+5)>1\gcd(n^5+5, (n+1)^5 + 5) > 1.

Note from above that dd must be a common divisor of both 50112n+2725650112n + 27256 and 1968751. Since 1968751 is a prime, if we have 50112n+27256≢0(mod1968751)50112n + 27256 \not \equiv 0 \pmod{1968751}, we must then have that d=1d=1.

Thus, we have 50112n1941495(mod1968751)50112n \equiv 1941495 \pmod{1968751}. We can solve this by finding the modular inverse of 50112. Since 50112 has small factors, we needn't apply Euclid's algorithm. We instead use inspection to find the inverse.

50112=26332950112 = 2^6 \cdot 3^3 \cdot 29. We will now check the value of each factor's inverse in (mod1968751)\pmod{1968751}.

2×984376=1968751+121984376(mod1968751)23984376÷4=246094(mod1968751)2624609421507325(mod1968751)\begin{aligned} 2 \times 984376 &= 1968751 + 1\\ \Rightarrow 2^{-1} &\equiv 984376 \pmod{1968751}\\ \Rightarrow 2^{-3} &\equiv 984376 \div 4 =246094 \pmod{1968751}\\ \Rightarrow 2^{-6} &\equiv 246094^2 \equiv 1507325 \pmod{1968751} \end{aligned}

Similarly, 331239584(mod1968751)3^{-3} \equiv 1239584 \pmod{1968751} and 2916788829^{-1} \equiv 67888. Thus, the modular inverse of 50112 is

50112126332911507325×1239584×678881731811(mod1968751)50112^{-1} \equiv 2^{-6} \cdot 3^{-3} \cdot 29^{-1} \equiv 1507325 \times 1239584 \times 67888 \equiv 1731811 \pmod{1968751}

Thus, we have

50112n1941495(mod1968751)n194195×1731811533360(mod1968751)\begin{aligned}50112n &\equiv 1941495 \pmod{1968751}\\ n &\equiv 194195 \times 1731811 \equiv 533360 \pmod{1968751} \end{aligned}

Therefore, when n533360(mod1968751)n \equiv 533360 \pmod{1968751}, n5+5n^5+5 and (n+1)5+5(n+1)^5+5 are not coprime.

Sharky Kesa - 3 years, 7 months ago

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This problem seems difficult than it looks... I have done half of the proof , just trll me no casework is required isnt it ? At first I tried to provethat by a little Modular Arith. but I am getting stuck in between...I am missing something basic...btw sharky you can arite the solution for shivams problem...This problem has made me crazy...@Sharky Kesa

A Former Brilliant Member - 3 years, 7 months ago

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I didn't use casework in my solution, mainly just divisibility and congruences. Also, because no one has yet given a solution, I will say that there are values of nn such that gcd(n5+5,(n+1)5+5)>1\gcd(n^5 + 5, (n+1)^5 + 5) > 1. @Chinmay Sangawadekar @Alex Spagnoletti

Sharky Kesa - 3 years, 7 months ago

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@Sharky Kesa Hint 2: You will want to know that 1968751 is a prime number if you wish to solve this. Also, use mainly divisibility laws (if dad \mid a and dbd \mid b, then d(ax+by)d \mid (ax+by)).

Sharky Kesa - 3 years, 7 months ago

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Solution of problem "extra partecipation problem -set 2: #2" n5+5n^5 + 5 and (n+1)5+5(n+1)^5 + 5 have no common divisors because seen that n5 n^5 and (n+1)5 (n+1)^5 are coprime for all positive integers nn if we add 5 to both they will remain coprime.

Alex Spagnoletti - 3 years, 7 months ago

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@Alex Spagnoletti That isn't true.Counterexample: gcd(1,3)=1gcd(1,3)=1 but gcd(1+5,3+5)=gcd(6,8)=21gcd(1+5,3+5)=gcd(6,8)=2\neq 1

Every pair of odd positive integers a,ba,b with gcd(a,b)=1gcd(a,b)=1 is a counterexample,as gcd(a+5,b+5)2gcd(a+5,b+5)\geq 2 as both a+5a+5 and b+5b+5 are even numbers.

Abdur Rehman Zahid - 3 years, 7 months ago

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@Abdur Rehman Zahid @Abdur Rehman Zahid, Yes but 3 3 is not 1+11 + 1 It must be nn and n+1n+1

Alex Spagnoletti - 3 years, 7 months ago

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This is not a correct solution. Just because two numbers are coprime doesnt mean that a constant added to both numbers will keep them both coprime.

Sharky Kesa - 3 years, 7 months ago

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@Sharky Kesa @Sharky Kesa It is not true for all numbers. But for n n and n+1 n+1 this is true. For example an even number can't be the gcd because they are one even and one odd. A 3k number can't be the gcd because if n is a 3k number 3k + 5 is not a 3k number and n and n+1 can be either one 3k and the other 3k+1( or 3k-1) or they can be both non 3k numbers. In this case adding 5 will make one 3k and the other not. For 5k numbers it's easy to see that if n is 5k, n+5 is also 5k but n+1 is not 5k and so it can't be the gcd. So it's the same for the other numbers and naturally this is the same with n5n^5 and (n+1)5(n+1)^5 seen that they are just n and (n+1) 5 times. We can also think to prove this absurdly. If n+5 and (n+1)+5 are not coprime this means that gcdn+5gcd|n+5 and gcdn+6gcd|n+6 but there is no number > 1 that can divide both a number and that number + 1 I think that this reasoning has no mistakes, but if I have made any mistakes please tell me because I can't find them.

Alex Spagnoletti - 3 years, 7 months ago

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@Alex Spagnoletti But we are referring to 5th power of consecutive numbers, not just consecutive numbers. The fifth power of consecutive numbers aren't consecutive.

Sharky Kesa - 3 years, 7 months ago

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@Sharky Kesa @Sharky Kesa It's right. But n5n^5 and (n+1)5( n+1)^5 are divisibile by n and (n+1) so it is the same. It is like if they are separated but they are always divisible by 2 consecutive numbers. So working modulo n I will have 0+50 + 5 and 1+51+5. So they are consecutive in this form. (And there is another time the absurd)

Alex Spagnoletti - 3 years, 7 months ago

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@Alex Spagnoletti No, your proof is incorrect. It has no mathematical rigour and is basically just an assumption you have incorrectly talked yourself into.

Sharky Kesa - 3 years, 7 months ago

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@Sharky Kesa Probably I have made some reasoning mistakes. If no one can find the solution can you post it?

Alex Spagnoletti - 3 years, 7 months ago

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@Alex Spagnoletti What you have done is a hasty generalisation. For example, you look at the odd numbers up to 13. 1 is a square number, 3 is a prime number, 5 is a prime number, 7 is a prime number, 9 is a square number, 11 is a prime number, 13 is a prime number. From this, you conclude that all odd numbers are either prime or square numbers, when 15 is an obvious counter-example.

Sharky Kesa - 3 years, 7 months ago

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Alternate Solution to Problem 5:
Recursion was my strategy.
Let S(N)=q=1nd2q1d2q S(N) = \sum\limits_{q = 1}^{n} d_{2q -1}\cdot d_{2q} where NN and did_i are as defined.
Let f(n) f(n) denote the number of ways we can find a suitable N of 2n2n digits, that satisfies S(N)0 mod 2 S(N) \equiv 0 \text{ mod } 2 .
We need to find a recurrence for f(n) f(n) .

Now consider N0 N_0, which is a number with 2n22n - 2 digits.

Now, if N0N_0 has S(N)S(N) even, then we need to find d2n1,d2nd_{2n - 1}, d_{2n} such that their product is even.
Note that the number of N0N_0 here is f(n1) f(n - 1) .
The number of ways to find such d2n1,d2nd_{2n - 1}, d_{2n} is f(1)=56f(1) = 56 , by simple casework.

If N0N_0 has S(N)S(N) odd, then we need to find d2n1,d2nd_{2n - 1}, d_{2n} such that their product is odd.
The number of the number of N0N_0 here is all the N0 N_0 not counted in the case above.
As there are 92n29^{2n - 2} possible numbers totally, and each number must fall in either of these categories, there are 92n2f(n1) 9^{2n - 2} - f(n - 1) such N0N_0 here.
The number of ways to find such d2n1,d2nd_{2n - 1}, d_{2n} is 92f(1)=259^2 - f(1) = 25 .
 

Then, f(n)=56f(n1)+25(92n2f(n1)) f(n) = 56\cdot f(n - 1) + 25 \cdot(9^{2n - 2} - f(n - 1)) for n>1n > 1 , and f(1)=56f(1) = 56
f(n)=31f(n1)+2592n2 \Rightarrow f(n) = 31 \cdot f(n - 1) + 25 \cdot 9^{2n - 2}

With some juggling,
f(n)=31f(n1)+81n3181n12 f(n) = 31 \cdot f(n - 1) + \dfrac{81^n - 31\cdot 81^{n - 1}}{2}

Noticing that, f(1)=56=81+312f(1) = 56 = \dfrac{81 + 31}{2} ,

We are motivated to try, f(n)=81n+31n2 f(n) = \dfrac{81^n + 31^n}{2} which satisfies the recurrence as well.

f(n)=81n+31n2 \therefore f(n) = \dfrac{81^n + 31^n}{2}

Ameya Daigavane - 3 years, 7 months ago

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Extra Participation Problem Set - 2 - #3

Let nn be an integer greater than 66. Prove that if n1n-1 and n+1n+1 are both prime then n2(n2+16)n^2(n^2+16) is divisible by 720720.

Bonus Prove that the converse is not true.

A Former Brilliant Member - 3 years, 7 months ago

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Being all the primes >5 >5 ends with 1,3,7 or 9 nn will end with 0,8 0, 8 or 2 2 because it must be in the middle of two primes. So 2n2|n and 3n3|n * for all nn means that 22n22^2|n^2 and 32n23^2|n^2. But seen that n is either 5k (if it ends with 0) or it is 2,8(mod10) \equiv 2, 8 \pmod{10} n2n^2 will be either 5k or 4(mod10) \equiv 4 \pmod{10}. So adding 16 will make n20(mod10)n^2 \equiv 0 \pmod {10} and so 5n5|n or in the case of n=5kn=5k it is already 0(mod10) \equiv 0 \pmod {10}. 720=24×32×5720=2^4×3^2×5 so n2(n2+16)n^2(n^2+16) can be divided in 2 cases:

1) if n2=22×32×5×?n^2=2^2×3^2×5×? and (n2+16)=22×?=>n2(n2+16)=24×32×5×? (n^2+16)=2^2×? => n^2(n^2+16) = 2^4×3^2×5×? and so it is divisible by 720.

2) if n2=22×32×?n^2=2^2×3^2×? and (n2+16)=22×5×?=>n2(n2+16)=24×32×5×?(n^2+16)=2^2×5×? => n^2(n^2+16) = 2^4×3^2×5×? and so it is divisible by 720.

  • the proof that 3n3|n is this: In a series of digits from 0 to 9 (we can think like if it is (mod10) \pmod{10} ) we can have the 0 the 3 the 6 and the 9 wich are divisible by 3. In this case nn can't be between 7 and 9 because 9 is not prime and neither between 1 and 3 because 3 is not prime. So it can only be the 0 because 1 is prime and also the 9 of an hypothetical previous series (wich can't be divisible by 3). While if are 2, 5 and 8 divisible by 3 this means that if I take 8 3n3|n and the same for 2. But I can't take 0 because it is between a 9 which is surely divisible by 3. If 1, 4 and 7 are divisible by 3 my nn could only be the 0 between my 9 and the 1 of a next series because 7 and 1 are divisible by 3. And also the next 0 is divisible by 3. In conclusion if we consider the last digit of any numbers we can always refer to this table and so all our n n are divisible by 3.

Alex Spagnoletti - 3 years, 7 months ago

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Correct! Here is my approach

Any prime number>3 is of the form 1,5(mod6)1,5 \pmod6. Thus nn is divisible by 66 take nn as 6m6m.

Thus it suffices to prove that m2(9m2+4)m^2(9m^2+4) is divisible by 55.

We eliminate the case m1,4(mod5)m \equiv 1,4 \pmod 5 because then n1n-1 and n+1n+1 will be divisible by 55 respectively(so it won't be prime then).

Checking the cases m0,2,3(mod5)m \equiv 0,2,3 \pmod5 we see that each of the cases satisfies.

Hence proved.

Bonus: Take n=1440n=1440 the expression is divisible by 720720 but 14411441 is not prime (11×131)(11 \times 131)

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member It is a shorter proof, I like it. However also n=258n=258 makes the expression dibisible by 720 720 but 259 259 is not prime. If nn is divisible by 6 and it ends with 8 , 4 or 0 the expression is satisfied also if (n+1)(n + 1) or (n1)(n - 1) are not prime.

Alex Spagnoletti - 3 years, 7 months ago

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@Alex Spagnoletti Yes. There are many ways to prove a counterexample for the bonus.

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member Yes. Now I have a question, (I'm new in this kind of constest and I don't know much), should I publish a problem here? Or it is just for those people who solve the regular problems?

Alex Spagnoletti - 3 years, 7 months ago

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@Alex Spagnoletti You should post a problem!

Akshat Sharda - 3 years, 7 months ago

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@Akshat Sharda Okok. Just give me the time to think about.

Alex Spagnoletti - 3 years, 7 months ago

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Problem 8

Let aa, bb, xx and yy be positive integers, with x>1x>1. Prove that if ax+bx=2ya^x + b^x = 2^y, then a=ba=b.


This question is based off AMO 2016 Q6, but a more general form.

Sharky Kesa - 3 years, 7 months ago

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Let aba \ge b, then we have axbxa^x \ge b^x or ax+bx2bxa^x+b^x \ge 2b^x

This means 2y1bx2^{y-1} \ge b^x

Now consider ba b \ge a then we have bxaxb^x \ge a^x or 2bxax+bx2b^x \ge a^x+b^x

This means bx2y1b^x \ge 2^{y-1}

Therefore bx=2y1b^{x} = 2^{y-1}

Similarly we can prove ax=2y1a^x=2^{y-1}.

Hence we have ax=bxa=ba^x=b^x \Longrightarrow a=b

How much will I get @Sharky Kesa

Department 8 - 3 years, 7 months ago

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You have proven the converse of the problem statement. In the first case, you have proven 2y1bx2^{y-1} \geq b^x if aba \geq b. In the second case, you have proven 2y1bx2^{y-1} \leq b^x if aba \leq b. From this, you are saying that if a=ba=b, then bx=2y1b^x = 2^{y-1}. Similarly, you are saying that ax=2y1a^x = 2^{y-1}, from which you get ax+bx=2ya^x + b^x = 2^y if a=ba=b (This is incorrect of course). In any case, you are trying to prove the converse of the problem statement.

Sharky Kesa - 3 years, 7 months ago

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Problem 3

Prove that

dnμ(n)={1 if n=10 if n>1\sum_{d \mid n}\mu(n)=\begin{cases} 1 \text{ if } n=1 \\ 0 \text{ if }n>1 \end{cases}

Not original.

This problem has been solved by Akshat Sharda

Julian Poon - 3 years, 7 months ago

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Solution to Problem 3

This is already a well known problem.

Let,

n=p1a1p2a2p3a3prarn=p_{1}^{a_1}\cdot p_{2}^{a_2}\cdot p_{3}^{a_3}\cdot \ldots \cdot p_{r}^{a_r}

As dnd|n,

d=p1b1p2b2p3b3prbrd=p_{1}^{b_1}\cdot p_{2}^{b_2}\cdot p_{3}^{b_3}\cdot \ldots \cdot p_{r}^{b_r}

Here, 0biai,i=1,2,,r0 ≤b_i ≤ a_i,\quad i=1,2,\ldots, r. If bi2,μ(d)=0b_i≥2, \quad \mu(d)=0. Therefore,

\begin{aligned} \displaystyle \sum_{d|n} \mu(d) & =\displaystyle \sum_{\substack{(b_1,b_2, \ldots , b_r) \\ b_i = 0 \text{ or } 1}} \mu( p_{1}^{b_1}\cdot p_{2}^{b_2}\cdot p_{3}^{b_3}\cdot \ldots \cdot p_{r}^{b_r}) \\ &=1 - \binom{r}{1} + \binom{r}{2} - \cdots + (-1)^{r} \binom {r}{r} \\ &= (1 - 1)^{r} \\ &= 0 \end{aligned}

Akshat Sharda - 3 years, 7 months ago

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Problem 4

Let {Xn}\{X_n\} and {Yn}\{Y_n\} denote 2 sequences of integers as follows:

X0=X1=1,Xn+1=Xn+2Xn1Y0=1,Y1=7,Yn+1=2Yn+3Yn1X_0=X_1=1, X_{n+1}=X_n+2X_{n-1} \\ Y_0=1, Y_1=7, Y_{n+1}=2Y_n+3Y_{n-1}

Prove that except for 1, there is no other term which occur in both sequences.

Not original.

This problem has been solved by Shivam Jadhav.

Akshat Sharda - 3 years, 7 months ago

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Answer to Problem 4: Using concept of linear recurrence relations we get xn=2n+1+(1)n3,yn=2(3)n+(1)n+1x_{n}=\frac{2^{n+1}+(-1)^{n}}{3},y_{n}=2(3)^{n}+(-1)^{n+1}

Now let xm=ynx_{m}=y_{n}

2(3n+12m)=3(1)m+(1)n2(3^{n+1}-2^{m})=3(-1)^{m}+(-1)^{n} RHS=4,4,2,2RHS=4,-4,2,-2 Therefore

(3n+12m)(3^{n+1}-2^{m}) can be equal to 2,2,1,12,-2,1,-1

But LHS=oddLHS=odd Since 2w=even,3q=odd2^{w}=even,3^{q}=odd and oddeven=oddodd-even=odd.

Therefore (3n+12m)=1,1 (3^{n+1}-2^{m})=1,-1 which is possible only for (m,n)=(0,1)(m,n)=(0,1) while taking the RHSRHS in consideration . Therefore only x0=y1x_{0}=y_{1} Hence proved .

Shivam Jadhav - 3 years, 7 months ago

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Extra Participation Problem - Set 1 - #1

Show that for every positive integer nn,

121n25n+1900n(4)n121^n-25^n+1900^n-(-4)^n is divisible by 20002000.

This problem has been solved by Ankit Kumar Jain and Akshat Sharda.

A Former Brilliant Member - 3 years, 7 months ago

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@Svatejas Shivakumar My solution is this:

What I will be doing is that I show that the expression is divisible by 125 and also by 16.

121n25n121^{n} -25^{n} is divisible by 121 - 25 = 96 and hence by 16. Also 1900n(4)n1900^{n} - (-4)^{n} is divisible by 1900 - (-4) = 1904 and hence by 16. So the expression is divisible by 16.

1900n25n1900^{n} - 25^{n} is divisible by 1900 - 25 = 1875 and hence by 125. Also 121n(4)n121^{n} - (-4)^{n} is divisible by 121 - (-4) = 125. So the whole expression is divisible by 125.

Hence the expression is divisible by 125×16125\times16...

As a shortcut you can straight away put n = 1 and check....

Ankit Kumar Jain - 3 years, 7 months ago

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121n25n+1900n(4)n(mod2)1n1n+000(mod2)121n25n+1900n(4)n(mod5)1n+0+0(4)0(mod5)n= odd1n+0+0(6)0(mod5)n= even121^n-25^n+1900^n-(-4)^n \pmod 2 \\ 1^n-1^n+0-0 \equiv 0 \pmod 2 \\ 121^n-25^n+1900^n-(-4)^n \pmod 5 \\ \underbrace{1^n+0+0-(-4) \equiv 0 \pmod 5}_{n=\text{ odd}} \\ \underbrace{1^n+0+0-(6) \equiv 0 \pmod 5}_{n=\text{ even}}

Therefore, 121n25n+1900n(4)n0(mod2000)121^n-25^n+1900^n-(-4)^n \equiv 0 \pmod {2000}

Akshat Sharda - 3 years, 7 months ago

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Can you please explain how you concluded that it is divisible by 2000(last step)?

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member This is where the Chinese Remainder theorem comes in.

Akshat Sharda - 3 years, 7 months ago

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Extra Participation Problem - Set 1 - #4

Prove that (n+1)(n+2)(2n1)(2n)(n+1)(n+2) \ldots (2n-1)(2n) is divisible by 2n2^{n}.

Harsh Shrivastava - 3 years, 7 months ago

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I think there's a relatively simpler argument.

For all positive integers nn, we have,

(n+1)(n+2)(2n1)(2n)=(2n)!n!=(2n)!!(2n1)!!n!=2nn!(2n1)!!n!=2n(2n1)!!(n+1)(n+2)\cdots (2n-1)(2n)=\frac{(2n)!}{n!}=\frac{(2n)!!(2n-1)!!}{n!}=\frac{2^n\cdot n!\cdot (2n-1)!!}{n!}=2^n (2n-1)!!

which is obviously divisible by 2n2^n for all nZ+n\in\Bbb Z^+ since (2n1)!!(2n-1)!! is an integer for all nZ+n\in\Bbb Z^+.

Notation: n!!n!! denotes the double factorial.

Prasun Biswas - 3 years, 7 months ago

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Solution to Extra Participation Problem-Set 1-#4

Here , we know that , (n+1)(n+2)...(2n1)(2n)0mod2(n+1)(n+2)...(2n-1)(2n) \equiv 0 \mod 2 , now multiplying the whole congruence by ,

2n12^{n-1} , we get , 2n1(n+1)(n+2)...(2n1)(2n)0mod2n2^{n-1}(n+1)(n+2)...(2n-1)(2n) \equiv 0 \mod 2^n , so from here , we know that , 2n2^{n}

does not divide 2n12^{n-1} but still remainder is 0 , so it is obvious that , 2n(n+1)(n+2)...(2n1)(2n)2^{n} \mid (n+1)(n+2)...(2n-1)(2n)

A Former Brilliant Member - 3 years, 7 months ago

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How do you conclude that since 2n∤2n12^n\not\mid 2^{n-1}, we must have 2ni=1n(n+i)2^n\mid \prod\limits_{i=1}^n (n+i) ? Consider a=24 , b=23 , c=2a=2^4~,~b=2^3~,~c=2, we have abca\mid bc and a∤ba\not\mid b but that doesn't imply aca\mid c.

Prasun Biswas - 3 years, 7 months ago

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Extra Participation Problem- Set 1 - #6

Prove that when nZ+,n1n \in \mathbb{Z^{+}}, n \not=1 , and 2n+n22^n + n^2 is prime, then n3(mod6)n\equiv 3\pmod{6} .

This problem is not original.

This problem has been solved by Svatejas Shivakumar.

Ralph James - 3 years, 7 months ago

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Solution to Extra Participation Problem - Set 1 - #6

If a numbers is a prime number >3, it is 1,5(mod6)\equiv 1,5 \pmod6

So, 2n+n21,5(mod6)2^n+n^2 \equiv 1,5 \pmod 6