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Question Number 5216    Answers: 1   Comments: 0

Let p_j represent the j−th prime number. Now, define the number n whose decimal representation is written out in terms of p_j (j∈N) in the following way: n=0.p_1 p_2 p_3 p_4 p_5 ...p_j p_(j+1) p_(j+2) ... or n=0.(2)(3)(5)(7)(11)...(521)(523)(541)... ⇒n=0.235711...521523541... Prove or disprove that n is irrational.

$${Let}\:{p}_{{j}} \:{represent}\:{the}\:{j}−{th}\:{prime}\:{number}. \\ $$$${Now},\:{define}\:{the}\:{number}\:{n}\:{whose} \\ $$$${decimal}\:{representation}\:{is}\:{written}\:{out} \\ $$$${in}\:{terms}\:{of}\:{p}_{{j}} \:\left({j}\in\mathbb{N}\right)\:{in}\:{the}\:{following} \\ $$$${way}: \\ $$$${n}=\mathrm{0}.{p}_{\mathrm{1}} {p}_{\mathrm{2}} {p}_{\mathrm{3}} {p}_{\mathrm{4}} {p}_{\mathrm{5}} ...{p}_{{j}} {p}_{{j}+\mathrm{1}} {p}_{{j}+\mathrm{2}} ... \\ $$$${or}\:{n}=\mathrm{0}.\left(\mathrm{2}\right)\left(\mathrm{3}\right)\left(\mathrm{5}\right)\left(\mathrm{7}\right)\left(\mathrm{11}\right)...\left(\mathrm{521}\right)\left(\mathrm{523}\right)\left(\mathrm{541}\right)... \\ $$$$\Rightarrow{n}=\mathrm{0}.\mathrm{235711}...\mathrm{521523541}... \\ $$$${Prove}\:{or}\:{disprove}\:{that}\:{n}\:{is}\:{irrational}. \\ $$$$ \\ $$$$ \\ $$

Question Number 5185    Answers: 1   Comments: 0

If 2^(x ) and 3^x are integers for some x∈R^+ , must x be an integer?

$${If}\:\:\mathrm{2}^{{x}\:} \:{and}\:\:\mathrm{3}^{{x}} \:{are}\:{integers}\:{for}\:{some} \\ $$$${x}\in\mathbb{R}^{+} ,\:{must}\:{x}\:{be}\:{an}\:{integer}?\: \\ $$

Question Number 5168    Answers: 0   Comments: 2

Find the value of 2023! (mod 2027).

$${Find}\:{the}\:{value}\:{of}\:\mathrm{2023}!\:\left({mod}\:\mathrm{2027}\right). \\ $$

Question Number 5165    Answers: 1   Comments: 1

what is (√(i+1)) ?

$${what}\:{is} \\ $$$$\sqrt{{i}+\mathrm{1}}\:\:\:\:? \\ $$

Question Number 5158    Answers: 1   Comments: 0

why 1 + (1/2) + (1/4) + (1/8) + ........... = 2

$${why} \\ $$$$\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:+\:...........\:=\:\mathrm{2} \\ $$$$ \\ $$

Question Number 5152    Answers: 0   Comments: 4

there is an ineterger a,b,c can there be an interger as a^n +b^(n+1) =c^(n+2) (n is a interger)

$${there}\:{is}\:{an}\:{ineterger}\:{a},{b},{c} \\ $$$${can}\:{there}\:{be}\:{an}\:{interger}\:{as}\: \\ $$$${a}^{{n}} +{b}^{{n}+\mathrm{1}} ={c}^{{n}+\mathrm{2}} \:\:\:\:\:\:\left({n}\:{is}\:{a}\:{interger}\right) \\ $$$$ \\ $$

Question Number 5144    Answers: 0   Comments: 5

Let n,j,q∈(Z^+ −{1}). Are there triples (n,j,q) such that the following conditions are satisfied altogether? (i) n=j^q (ii)n^2 =j^2 +q^2 −−−−−−−−−−−−−−−−−−−−−− Suppose then that condition (ii) above is replaced by the following condition: (iii) n^2 =rj^2 +q^2 where r∈(Z−{0,1}) What solutions (n,j,q) exist in this case?

$${Let}\:{n},{j},{q}\in\left(\mathbb{Z}^{+} −\left\{\mathrm{1}\right\}\right).\:{Are}\:{there}\: \\ $$$${triples}\:\left({n},{j},{q}\right)\:{such}\:{that}\:{the}\:{following} \\ $$$${conditions}\:{are}\:{satisfied}\:{altogether}? \\ $$$$\left({i}\right)\:{n}={j}^{{q}} \:\:\:\: \\ $$$$\left({ii}\right){n}^{\mathrm{2}} ={j}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Suppose}\:{then}\:{that}\:{condition}\:\left({ii}\right)\:{above} \\ $$$${is}\:{replaced}\:{by}\:{the}\:{following}\:{condition}: \\ $$$$ \\ $$$$\left({iii}\right)\:{n}^{\mathrm{2}} ={rj}^{\mathrm{2}} +{q}^{\mathrm{2}} \:{where}\:{r}\in\left(\mathbb{Z}−\left\{\mathrm{0},\mathrm{1}\right\}\right) \\ $$$$ \\ $$$${What}\:{solutions}\:\left({n},{j},{q}\right)\:{exist}\:{in}\:{this}\:{case}? \\ $$$$ \\ $$

Question Number 5125    Answers: 2   Comments: 3

Question Number 5082    Answers: 2   Comments: 0

a^2 −b^2 =?

$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =? \\ $$

Question Number 4760    Answers: 0   Comments: 2

the number 27000001 has 4 prime factors. find thier sum

$${the}\:{number}\:\mathrm{27000001} \\ $$$${has}\:\mathrm{4}\:{prime}\:{factors}. \\ $$$${find}\:{thier}\:{sum} \\ $$

Question Number 4540    Answers: 1   Comments: 2

Let us define the positive number n with four digits a,b,c and d such that n=abcd with a,b,c,d∈Z, 1≤a≤9, 0≤b≤9, 0≤c≤9 and 0≤d≤9. Let us then say that a cool number is a four digit number, say n, such that the two digit numbers written as ab and cd are given by ab=r×s and cd=(r−1)×(s+1) for some non−negative integers r and s, r≠s. For example, 8081 has a=8,b=0 and 80=10×8= while c=8,d=1 and 81=9×9=(10−1)(8+1). So, for n=8081, r=10 while s=8. How many n, for 1000≤n≤9999, are cool? For n∈[1000,9999],n∈Z, how many n exist so that ab=r×s and cd=r×s+1? Call such n warm numbers.

$${Let}\:{us}\:{define}\:{the}\:{positive}\:{number}\:{n}\:{with}\:{four} \\ $$$${digits}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:{and}\:\boldsymbol{\mathrm{d}}\:{such}\:{that}\:{n}=\boldsymbol{\mathrm{abcd}} \\ $$$${with}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}},\boldsymbol{\mathrm{d}}\in\mathbb{Z},\:\mathrm{1}\leqslant\boldsymbol{\mathrm{a}}\leqslant\mathrm{9},\:\mathrm{0}\leqslant\boldsymbol{\mathrm{b}}\leqslant\mathrm{9}, \\ $$$$\mathrm{0}\leqslant\boldsymbol{\mathrm{c}}\leqslant\mathrm{9}\:{and}\:\mathrm{0}\leqslant\boldsymbol{\mathrm{d}}\leqslant\mathrm{9}.\:{Let}\:{us}\:{then}\:{say} \\ $$$${that}\:{a}\:{cool}\:{number}\:{is}\:{a}\:{four}\:{digit}\:{number}, \\ $$$${say}\:{n},\:{such}\:{that}\:{the}\:{two}\:{digit}\:{numbers}\:{written}\:{as} \\ $$$$\boldsymbol{\mathrm{ab}}\:{and}\:\boldsymbol{\mathrm{cd}}\:{are}\:{given}\:{by}\:\boldsymbol{\mathrm{ab}}={r}×{s}\:{and} \\ $$$$\boldsymbol{\mathrm{cd}}=\left({r}−\mathrm{1}\right)×\left({s}+\mathrm{1}\right)\:{for}\:{some}\:{non}−{negative}\:{integers} \\ $$$${r}\:{and}\:{s},\:{r}\neq{s}.\:{For}\:{example},\:\mathrm{8081}\:{has}\: \\ $$$$\boldsymbol{\mathrm{a}}=\mathrm{8},\boldsymbol{\mathrm{b}}=\mathrm{0}\:{and}\:\mathrm{80}=\mathrm{10}×\mathrm{8}=\:{while}\: \\ $$$$\boldsymbol{\mathrm{c}}=\mathrm{8},\boldsymbol{\mathrm{d}}=\mathrm{1}\:{and}\:\mathrm{81}=\mathrm{9}×\mathrm{9}=\left(\mathrm{10}−\mathrm{1}\right)\left(\mathrm{8}+\mathrm{1}\right). \\ $$$${So},\:{for}\:{n}=\mathrm{8081},\:{r}=\mathrm{10}\:{while}\:{s}=\mathrm{8}. \\ $$$${How}\:{many}\:{n},\:{for}\:\mathrm{1000}\leqslant{n}\leqslant\mathrm{9999},\:{are}\:{cool}? \\ $$$$ \\ $$$${For}\:{n}\in\left[\mathrm{1000},\mathrm{9999}\right],{n}\in\mathbb{Z},\:{how}\:{many}\:{n}\:{exist} \\ $$$${so}\:{that}\:\boldsymbol{\mathrm{ab}}={r}×{s}\:{and}\:\boldsymbol{\mathrm{cd}}={r}×{s}+\mathrm{1}?\:{Call} \\ $$$${such}\:{n}\:{warm}\:{numbers}.\:\: \\ $$$$ \\ $$

Question Number 4362    Answers: 0   Comments: 1

Determine integers x,y,z satisfying: ax^b +by^c =cz^a a,b,c are fixed integers.

$$\mathrm{Determine}\:\mathrm{integers}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{satisfying}: \\ $$$$\mathrm{ax}^{\mathrm{b}} +\mathrm{by}^{\mathrm{c}} =\mathrm{cz}^{\mathrm{a}} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{integers}. \\ $$

Question Number 4358    Answers: 0   Comments: 1

Determine integer solution of bx^a +ay^b =cz^c a,b,c are fixed integers.

$$\mathrm{Determine}\:\mathrm{integer}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\mathrm{bx}^{\mathrm{a}} +\mathrm{ay}^{\mathrm{b}} =\mathrm{cz}^{\mathrm{c}} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{integers}. \\ $$

Question Number 4242    Answers: 0   Comments: 11

Find that value of 2^2^(2∙∙∙) (continued power of 2) using analytical continuation.

$$\mathrm{Find}\:\mathrm{that}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{2}^{\mathrm{2}^{\mathrm{2}\centerdot\centerdot\centerdot} } \:\:\left(\mathrm{continued}\:\mathrm{power}\:\mathrm{of}\:\mathrm{2}\right) \\ $$$$\mathrm{using}\:\mathrm{analytical}\:\mathrm{continuation}. \\ $$

Question Number 4230    Answers: 2   Comments: 1

Solve for +ve integers >0. x^2 +y^4 =z^6

$$\mathrm{Solve}\:\mathrm{for}\:+\mathrm{ve}\:\mathrm{integers}\:>\mathrm{0}. \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{4}} ={z}^{\mathrm{6}} \\ $$

Question Number 4203    Answers: 2   Comments: 2

I have no formal background in number theory, but I′m curious of how to find positive integer solutions (x,y,z) to the equation x^n +y^n =z^n for n∈Z^− . Fermat′s last theorem led me to this. Tell me about the cases of n=−1,n=−2 and n=−3.

$${I}\:{have}\:{no}\:{formal}\:{background}\:{in}\: \\ $$$${number}\:{theory},\:{but}\:{I}'{m}\:{curious} \\ $$$${of}\:{how}\:{to}\:{find}\:{positive}\:{integer}\:{solutions}\: \\ $$$$\left({x},{y},{z}\right)\:{to}\:{the}\:{equation}\:{x}^{{n}} +{y}^{{n}} ={z}^{{n}} \:{for}\: \\ $$$${n}\in\mathbb{Z}^{−} .\:{Fermat}'{s}\:{last}\:{theorem}\:{led} \\ $$$${me}\:{to}\:{this}.\:{Tell}\:{me}\:{about}\:{the}\:{cases} \\ $$$${of}\:{n}=−\mathrm{1},{n}=−\mathrm{2}\:{and}\:{n}=−\mathrm{3}. \\ $$

Question Number 4199    Answers: 1   Comments: 0

Let p∈P and m∈Z^+ . Find (p,m) such that p^(m−1) (p−1)=146410.

$${Let}\:{p}\in\mathbb{P}\:{and}\:{m}\in\mathbb{Z}^{+} . \\ $$$${Find}\:\left({p},{m}\right)\:{such}\:{that}\:{p}^{{m}−\mathrm{1}} \left({p}−\mathrm{1}\right)=\mathrm{146410}. \\ $$

Question Number 3685    Answers: 1   Comments: 0

let p_i be the i^(th) prime Does the fillowing sum converge: Σ_(i=1) ^∞ (p_i /p_(i+1) ) (p_1 =2, p_2 =3, p_3 =5, ...)

$$\mathrm{let}\:{p}_{{i}} \:\mathrm{be}\:\mathrm{the}\:{i}^{\mathrm{th}} \:\mathrm{prime} \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{fillowing}\:\mathrm{sum}\:\mathrm{converge}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{p}_{{i}} }{{p}_{{i}+\mathrm{1}} } \\ $$$$\left({p}_{\mathrm{1}} =\mathrm{2},\:{p}_{\mathrm{2}} =\mathrm{3},\:{p}_{\mathrm{3}} =\mathrm{5},\:...\right) \\ $$

Question Number 3659    Answers: 0   Comments: 6

p_i = ith prime let: ρ=p_n −p_(n−1) is: lim_(n→∞) ρ=∞?

$${p}_{{i}} \:=\:{i}\mathrm{th}\:\mathrm{prime} \\ $$$$\mathrm{let}:\:\:\:\:\:\:\rho={p}_{{n}} −{p}_{{n}−\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{is}: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\rho=\infty? \\ $$

Question Number 3630    Answers: 0   Comments: 0

Prove that sum of all prime numbers p such that n≤p≤2^k n is ≥2^k n. Σ_(n≤p≤2^k n) p ≥ 2^k n (p−prime)

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{prime}\:\mathrm{numbers}\:{p} \\ $$$$\mathrm{such}\:\mathrm{that}\:{n}\leqslant{p}\leqslant\mathrm{2}^{{k}} {n}\:\mathrm{is}\:\geqslant\mathrm{2}^{{k}} {n}. \\ $$$$\underset{{n}\leqslant{p}\leqslant\mathrm{2}^{{k}} {n}} {\sum}{p}\:\:\geqslant\:\mathrm{2}^{{k}} {n}\:\:\:\:\:\left({p}−{prime}\right) \\ $$

Question Number 3644    Answers: 1   Comments: 1

An hexagon of unit side is drawn on plane. Draw a square having the same area as the hexagon using only unmarked ruler and compass. What if an n−gon with unit edges is given? Is it always possible to draw a square of the same area as n−gon using ruler and compass.

$$\mathrm{An}\:\mathrm{hexagon}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{side}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{on}\:\mathrm{plane}. \\ $$$$\mathrm{Draw}\:\mathrm{a}\:\mathrm{square}\:\mathrm{having}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area}\:\mathrm{as}\:\mathrm{the} \\ $$$$\mathrm{hexagon}\:\mathrm{using}\:\mathrm{only}\:\mathrm{unmarked}\:\mathrm{ruler}\:\mathrm{and}\: \\ $$$$\mathrm{compass}. \\ $$$$\mathrm{What}\:\mathrm{if}\:\mathrm{an}\:{n}−\mathrm{gon}\:\mathrm{with}\:\mathrm{unit}\:\mathrm{edges}\:\mathrm{is}\:\mathrm{given}? \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{always}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area}\:\mathrm{as}\:{n}−\mathrm{gon}\:\mathrm{using}\:\mathrm{ruler} \\ $$$$\mathrm{and}\:\mathrm{compass}. \\ $$

Question Number 3595    Answers: 0   Comments: 10

I just thought of something I am curious in figuring out. All integer numbers can be made up by prime factors. That is: n=p_1 ×p_2 ×...×p_i n∈Z p_k ∈P Are there an inifinite number of numbers that are the sum of prime numbers? That is: P=p_1 +p_2 +...+p_i P,p_k ∈P For example: 2+3=5 2+3+3+5=13 etc. Are all of these special primes odd? What else can we work out?

$$\mathrm{I}\:\mathrm{just}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{something}\:\mathrm{I}\:\mathrm{am}\:\mathrm{curious} \\ $$$$\mathrm{in}\:\mathrm{figuring}\:\mathrm{out}. \\ $$$$ \\ $$$$\mathrm{All}\:\mathrm{integer}\:\mathrm{numbers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{up}\:\mathrm{by} \\ $$$${prime}\:{factors}.\:\mathrm{That}\:\mathrm{is}: \\ $$$${n}={p}_{\mathrm{1}} ×{p}_{\mathrm{2}} ×...×{p}_{{i}} \\ $$$${n}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:{p}_{{k}} \in\mathbb{P} \\ $$$$ \\ $$$$\mathrm{Are}\:\mathrm{there}\:\mathrm{an}\:\mathrm{inifinite}\:\mathrm{number}\:\mathrm{of}\:\mathrm{numbers} \\ $$$$\mathrm{that}\:\mathrm{are}\:\mathrm{the}\:{sum}\:\mathrm{of}\:{prime}\:{numbers}? \\ $$$$\mathrm{That}\:\mathrm{is}: \\ $$$${P}={p}_{\mathrm{1}} +{p}_{\mathrm{2}} +...+{p}_{{i}} \\ $$$${P},{p}_{{k}} \in\mathbb{P} \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{example}: \\ $$$$\mathrm{2}+\mathrm{3}=\mathrm{5} \\ $$$$\mathrm{2}+\mathrm{3}+\mathrm{3}+\mathrm{5}=\mathrm{13} \\ $$$${etc}. \\ $$$$ \\ $$$$\mathrm{Are}\:\mathrm{all}\:\mathrm{of}\:\mathrm{these}\:\mathrm{special}\:\mathrm{primes}\:\mathrm{odd}? \\ $$$$\mathrm{What}\:\mathrm{else}\:\mathrm{can}\:\mathrm{we}\:\mathrm{work}\:\mathrm{out}? \\ $$

Question Number 3519    Answers: 1   Comments: 5

n is a number such that regular n−gon is possible with straightedge and compass only. ∗Write first thirty values of n. ∗What are other properties of such numbers ? ∗If values of n are arranged in order, what is the formula for generating Nth number?

$${n}\:{is}\:{a}\:{number}\:{such}\:{that}\:{regular}\:\:{n}−{gon}\:{is} \\ $$$${possible}\:{with}\:{straightedge}\:{and}\:\:{compass}\:{only}. \\ $$$$\ast{Write}\:{first}\:{thirty}\:{values}\:{of}\:{n}. \\ $$$$\ast{What}\:{are}\:{other}\:{properties}\:{of}\:{such}\:{numbers}\:? \\ $$$$\ast{If}\:{values}\:{of}\:{n}\:{are}\:{arranged}\:{in}\:{order},\:{what}\:{is} \\ $$$${the}\:{formula}\:{for}\:{generating}\:{Nth}\:{number}? \\ $$

Question Number 3296    Answers: 1   Comments: 3

If n,p,q∈Z^+ and p and q are coprimes, then prove that HCF of (n^p −1) and (n^q −1) is (n−1). Assume n>1.

$$\mathrm{If}\:{n},{p},{q}\in\mathbb{Z}^{+} \:\mathrm{and}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{coprimes}, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{HCF}\:\mathrm{of}\:\left({n}^{{p}} −\mathrm{1}\right)\:\mathrm{and}\:\left({n}^{{q}} −\mathrm{1}\right)\:\mathrm{is}\:\left({n}−\mathrm{1}\right). \\ $$$$\mathrm{Assume}\:{n}>\mathrm{1}. \\ $$

Question Number 3205    Answers: 1   Comments: 9

Suggest minimum number of weights ,two peices of each, to weigh upto at least 60 kg(in whole kg′s) in a common balance.

$$\mathcal{S}{uggest}\:{minimum}\:{number}\:{of}\:\:{weights}\:,{two}\:{peices}\:{of}\:{each},\: \\ $$$${to}\:{weigh}\:{upto}\:{at}\:{least}\:\mathrm{60}\:{kg}\left({in}\:{whole}\:{kg}'{s}\right)\:{in}\:{a}\:{common} \\ $$$${balance}. \\ $$

Question Number 3172    Answers: 1   Comments: 1

How many different clock−type dials can be made containing first n natual numbers with the property that sum of any two numbers of consecutive positions be a prime number. N={1,2,3,...}

$$\mathcal{H}{ow}\:{many}\:{different}\:\:{clock}−{type}\:{dials}\:{can}\:{be}\:{made}\: \\ $$$${containing}\:{first}\:{n}\:{natual}\:{numbers}\:{with}\:{the}\:{property} \\ $$$${that}\:{sum}\:{of}\:\:{any}\:{two}\:{numbers}\:{of}\:{consecutive}\:{positions}\:{be} \\ $$$${a}\:{prime}\:{number}. \\ $$$$\mathbb{N}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\right\} \\ $$

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