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Question Number 92687    Answers: 0   Comments: 0

In a Gregorian calendar, a year finishing with 00 is a leap year if only it′s vintage is divisible by 400. Also, the 1^(st) January 1900 was a Monday. 1\ Show that a year with the vintage finishing with 00 cannot begin on a Sunday 2\ Show that for a person born between 1900−2071, his 28 anniversary will occur on the same day of birth.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{Gregorian}\:\mathrm{calendar},\:\mathrm{a}\:\mathrm{year}\:\mathrm{finishing}\:\mathrm{with} \\ $$$$\mathrm{00}\:\mathrm{is}\:\mathrm{a}\:\mathrm{leap}\:\mathrm{year}\:\mathrm{if}\:\mathrm{only}\:\mathrm{it}'\mathrm{s}\:\mathrm{vintage}\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{400}.\:\mathrm{Also},\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{January}\:\mathrm{1900}\:\mathrm{was}\:\mathrm{a} \\ $$$$\mathrm{Monday}. \\ $$$$\mathrm{1}\backslash\:\mathrm{Show}\:\mathrm{that}\:\mathrm{a}\:\mathrm{year}\:\mathrm{with}\:\mathrm{the}\:\mathrm{vintage}\:\mathrm{finishing}\:\mathrm{with} \\ $$$$\mathrm{00}\:\mathrm{cannot}\:\mathrm{begin}\:\mathrm{on}\:\mathrm{a}\:\mathrm{Sunday} \\ $$$$\mathrm{2}\backslash\:\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{a}\:\mathrm{person}\:\mathrm{born}\:\mathrm{between}\:\mathrm{1900}−\mathrm{2071}, \\ $$$$\mathrm{his}\:\mathrm{28}\:\mathrm{anniversary}\:\mathrm{will}\:\mathrm{occur}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{day}\:\mathrm{of}\:\mathrm{birth}. \\ $$

Question Number 92673    Answers: 1   Comments: 2

Show that if 3 prime numbers, all greater than 3, form an arithmetic progression then the common difference of the progression is divisible by 6.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{3}\:\mathrm{prime}\:\mathrm{numbers},\:\mathrm{all}\:\mathrm{greater} \\ $$$$\mathrm{than}\:\mathrm{3},\:\mathrm{form}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}\:\mathrm{then}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{progression}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{6}. \\ $$

Question Number 90801    Answers: 0   Comments: 3

what are the next two number of this series 4,7,9,2,5,6,3,8 ?

$${what}\:{are}\:{the}\:{next}\:{two}\:{number}\: \\ $$$${of}\:{this}\:{series}\:\mathrm{4},\mathrm{7},\mathrm{9},\mathrm{2},\mathrm{5},\mathrm{6},\mathrm{3},\mathrm{8}\:? \\ $$

Question Number 90233    Answers: 0   Comments: 5

3,4,12,39,103,x (a) 164 (b) 170 (c) 172 (d) 228

$$\mathrm{3},\mathrm{4},\mathrm{12},\mathrm{39},\mathrm{103},\mathrm{x}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{164} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{170} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{172} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{228} \\ $$

Question Number 90200    Answers: 0   Comments: 7

26≡R_1 [37] 1 ≡R_2 [3] 2≡R_3 [5] Find R_1 , R_2 and R_3

$$\mathrm{26}\equiv\mathrm{R}_{\mathrm{1}} \left[\mathrm{37}\right] \\ $$$$\mathrm{1}\:\:\equiv\mathrm{R}_{\mathrm{2}} \left[\mathrm{3}\right] \\ $$$$\mathrm{2}\equiv\mathrm{R}_{\mathrm{3}} \left[\mathrm{5}\right] \\ $$$$\mathrm{Find}\:\mathrm{R}_{\mathrm{1}} ,\:\mathrm{R}_{\mathrm{2}} \mathrm{and}\:\mathrm{R}_{\mathrm{3}} \\ $$

Question Number 88943    Answers: 0   Comments: 2

Question Number 88940    Answers: 1   Comments: 0

Question Number 88690    Answers: 1   Comments: 0

Question Number 85555    Answers: 1   Comments: 0

2x^2 +5x+7=0

$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\mathrm{0} \\ $$

Question Number 79462    Answers: 1   Comments: 0

prove p⇒q and ∼q⇒∼p are logicaly equivalent with out truth table

$${prove}\:{p}\Rightarrow{q}\:{and}\:\sim{q}\Rightarrow\sim{p}\:{are}\:{logicaly}\: \\ $$$${equivalent}\:{with}\:{out}\:{truth}\:{table} \\ $$$$ \\ $$

Question Number 78671    Answers: 1   Comments: 0

prove p⇒q and negetion of q⇒negation of p

$${prove}\:{p}\Rightarrow{q}\:{and}\:{negetion}\:{of}\:{q}\Rightarrow{negation}\:{of}\:{p} \\ $$

Question Number 75255    Answers: 0   Comments: 0

Question Number 75254    Answers: 1   Comments: 0

Question Number 75257    Answers: 1   Comments: 0

Let A={2,4,6,7,8,9} B={1,3,5,6,10} and C={x:3x+6=0 or 2x+6=0}.Find a. A∪B. b. is(A∪B)∪C=A∪(B∪C)?

$${Let}\:{A}=\left\{\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\} \\ $$$${B}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{10}\right\}\:{and} \\ $$$${C}=\left\{{x}:\mathrm{3}{x}+\mathrm{6}=\mathrm{0}\:{or}\:\mathrm{2}{x}+\mathrm{6}=\mathrm{0}\right\}.{Find} \\ $$$${a}.\:{A}\cup{B}. \\ $$$${b}.\:{is}\left({A}\cup{B}\right)\cup{C}={A}\cup\left({B}\cup{C}\right)? \\ $$

Question Number 74246    Answers: 2   Comments: 0

Question Number 71849    Answers: 1   Comments: 0

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Question Number 70742    Answers: 0   Comments: 1

Prove that f(x) = x − a cos (x) − b has at least one real root for ∀a,b ∈ R

$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)\:=\:{x}\:−\:{a}\:\mathrm{cos}\:\left({x}\right)\:−\:{b} \\ $$$$\mathrm{has}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}\:\mathrm{for}\:\forall{a},{b}\:\in\:\mathbb{R} \\ $$

Question Number 70168    Answers: 1   Comments: 3

Question Number 69116    Answers: 0   Comments: 0

Here are three propositions: − This sentence has exactly six words −There are two wrong propositions −The two previous sentences are correct Among that propositions ,how many are wrong? list them!

$$\:{Here}\:\:{are}\:{three}\:{propositions}:\: \\ $$$$−\:{This}\:{sentence}\:{has}\:{exactly}\:{six}\:{words} \\ $$$$−{There}\:{are}\:{two}\:{wrong}\:{propositions} \\ $$$$−{The}\:{two}\:{previous}\:{sentences}\:{are}\:{correct} \\ $$$$ \\ $$$${Among}\:{that}\:{propositions}\:,{how}\:{many}\:{are}\:{wrong}?\:{list}\:{them}! \\ $$

Question Number 68695    Answers: 1   Comments: 0

pour 1<k<n montrer que k(n+1−k)<(n+1/2)^2

$${pour}\:\mathrm{1}<{k}<{n}\:\:\:\:\:{montrer}\:{que} \\ $$$${k}\left({n}+\mathrm{1}−{k}\right)<\left({n}+\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} \\ $$

Question Number 67651    Answers: 0   Comments: 0

1)Let consider S= Σ_(n=0) ^∞ n and T=Σ_(n=0) ^∞ (−1)^(n+1) n We know that ∀ x∈]−1;1] Σ_(n=0) ^∞ (−x)^n =(1/(1+x)) , then after derivating (1/((1+x)^2 ))=Σ_(n=1) ^∞ (−1)^(n+1) nx^(n−1) for x=1 ,we get T=(1/4) Now let ascertain something T=Σ_(n=0) ^∞ (−1)^(2n+2) (2n+1) +Σ_(n=0) ^∞ (−1)^(2n+1) (2n) =Σ_(n=0) ^∞ (2n+1) −2S (•) knowing that S=Σ_(n=0) ^∞ (2n)+Σ_(n=0) ^∞ (2n+1) So Σ_(n=0) ^∞ (2n+1)= S−2S When replacing that value in (•) we get T=(S−2S )−2S If i conclude that T=−3S and finally find S=−(T/3)=−(1/(12)) where will the mistake be? 2)Let consider K=1+((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +......+... ((2020)/(2019))K=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +.... K−1=((2020)/(2019))+(((2020)/(2019)))^2 +(((2020)/(2019)))^3 +... So ((2020)/(2019))K=K−1 Then K=−2019 Where is the error? 3)Let consider n an integer We have 0=n−n =n+(−n)=n+(−n)^(2×(1/2)) =n+[(−n)^2 ]^(1/2) =n+ (√n^2 ) = n+n=2n So << all integer are null : 0 is the only integer>> Where is the error? 4) let consider n an integer different of zero and f(n)=nln(n) we have (df/dn)=ln(n)+1 (•) Likewise f(n)=ln(n^n ) and we know that n^n =n×n×n×......×n (n times) So f(n)=ln(n)+ln(n)+......+ln(n) (n times) Now we have (df/dn)=(1/n)+(1/n)+....+(1/n) (n times) So (df/( dn))=1 (••) Relation (•) and (••) give ln(n)+1=1 then ln(n)=0 ⇒ n=1 << The logarithm of all n≥1 is null : There is no integer big than 1 >> Where is the error? 5) let consider x=0,999999999....... we ascertain that 10x=9,999999999...... then 10x=9+0,999999999.... So 10x=9+x finally x=1 << 0.9999999999999999 ..... is and integer >> Is there any error? 6)let consider a=((26666666666666666)/(66666666666666665)) b=((999999999999999999999995)/(199999999999999999999999)) In the way to cancel , if i just remove one common figure to the numerator and to the denominator And i find a=(2/5) and b=(5/1) Will it be wrong? if no ,explain!

$$\left.\mathrm{1}\right){Let}\:{consider}\:\:{S}=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{n}\:\:\:\:\:{and}\:\:{T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {n} \\ $$$$\left.{W}\left.{e}\:{know}\:{that}\:\:\forall\:{x}\in\right]−\mathrm{1};\mathrm{1}\right] \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:,\:\: \\ $$$${then}\:{after}\:{derivating}\: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{nx}^{{n}−\mathrm{1}} \\ $$$${for}\:\:{x}=\mathrm{1}\:,{we}\:{get}\:\:\:{T}=\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$${Now}\:\:{let}\:{ascertain}\:{something} \\ $$$${T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)\:+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}{n}\right) \\ $$$$\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)\:−\mathrm{2}{S}\:\:\:\:\:\:\:\:\:\:\:\:\left(\bullet\right) \\ $$$${knowing}\:{that}\:{S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}\right)+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${So}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)=\:{S}−\mathrm{2}{S}\: \\ $$$${When}\:{replacing}\:{that}\:{value}\:{in}\:\:\left(\bullet\right) \\ $$$${we}\:{get}\:\: \\ $$$${T}=\left({S}−\mathrm{2}{S}\:\right)−\mathrm{2}{S} \\ $$$$\: \\ $$$${If}\:{i}\:{conclude}\:{that}\:\:{T}=−\mathrm{3}{S}\:\: \\ $$$${and}\:{finally}\:\:{find}\:\:{S}=−\frac{{T}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\: \\ $$$${where}\:{will}\:{the}\:{mistake}\:{be}? \\ $$$$\left.\:\mathrm{2}\right){Let}\:{consider}\:{K}=\mathrm{1}+\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +......+... \\ $$$$\frac{\mathrm{2020}}{\mathrm{2019}}{K}=\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +.... \\ $$$${K}−\mathrm{1}=\frac{\mathrm{2020}}{\mathrm{2019}}+\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right)^{\mathrm{3}} +... \\ $$$${So}\:\:\frac{\mathrm{2020}}{\mathrm{2019}}{K}={K}−\mathrm{1}\: \\ $$$${Then}\:\:{K}=−\mathrm{2019} \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{3}\right){Let}\:{consider}\:{n}\:{an}\:\:{integer}\: \\ $$$${We}\:{have} \\ $$$$\mathrm{0}={n}−{n} \\ $$$$\:\:={n}+\left(−{n}\right)={n}+\left(−{n}\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:={n}+\left[\left(−{n}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} ={n}+\:\sqrt{{n}^{\mathrm{2}} }\:=\:{n}+{n}=\mathrm{2}{n} \\ $$$${So}\:<<\:{all}\:\:{integer}\:\:\:{are}\:{null}\:\::\:\:\mathrm{0}\:{is}\:{the}\:\:{only}\:{integer}>> \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{4}\right)\:\:\:{let}\:{consider}\:{n}\:{an}\:{integer}\:{different}\:{of}\:{zero}\:{and}\:\:{f}\left({n}\right)={nln}\left({n}\right) \\ $$$${we}\:{have}\:\:\frac{{df}}{{dn}}={ln}\left({n}\right)+\mathrm{1}\:\:\:\:\left(\bullet\right) \\ $$$${Likewise}\:{f}\left({n}\right)={ln}\left({n}^{{n}} \right)\:\:\:{and}\:{we}\:{know}\:{that} \\ $$$${n}^{{n}} ={n}×{n}×{n}×......×{n}\:\:\left({n}\:{times}\right) \\ $$$${So}\:\:{f}\left({n}\right)={ln}\left({n}\right)+{ln}\left({n}\right)+......+{ln}\left({n}\right)\:\:\:\:\:\:\left({n}\:{times}\right) \\ $$$${Now}\:{we}\:{have}\:\:\frac{{df}}{{dn}}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}+....+\frac{\mathrm{1}}{{n}}\:\:\:\:\:\:\left({n}\:\:{times}\right) \\ $$$${So}\:\:\:\frac{{df}}{\:{dn}}=\mathrm{1}\:\:\:\:\:\left(\bullet\bullet\right) \\ $$$${Relation}\:\left(\bullet\right)\:\:{and}\:\:\:\:\left(\bullet\bullet\right)\:\:{give}\:\:\:\: \\ $$$${ln}\left({n}\right)+\mathrm{1}=\mathrm{1}\:\:\:{then}\:\:{ln}\left({n}\right)=\mathrm{0}\:\Rightarrow\:{n}=\mathrm{1} \\ $$$$<<\:{The}\:{logarithm}\:{of}\:\:{all}\:{n}\geqslant\mathrm{1}\:{is}\:{null}\::\:{There}\:{is}\:{no}\:{integer}\:{big}\:{than}\:\mathrm{1}\:>> \\ $$$${Where}\:{is}\:{the}\:{error}? \\ $$$$\left.\mathrm{5}\right)\:{let}\:{consider}\:\:{x}=\mathrm{0},\mathrm{999999999}....... \\ $$$${we}\:{ascertain}\:{that}\: \\ $$$$\mathrm{10}{x}=\mathrm{9},\mathrm{999999999}...... \\ $$$${then}\:\:\mathrm{10}{x}=\mathrm{9}+\mathrm{0},\mathrm{999999999}.... \\ $$$${So}\:\:\mathrm{10}{x}=\mathrm{9}+{x}\:\: \\ $$$${finally}\:\:{x}=\mathrm{1} \\ $$$$<<\:\mathrm{0}.\mathrm{9999999999999999}\:.....\:\:{is}\:\:{and}\:{integer}\:>> \\ $$$${Is}\:{there}\:{any}\:{error}? \\ $$$$\left.\mathrm{6}\right){let}\:{consider}\:\:{a}=\frac{\mathrm{26666666666666666}}{\mathrm{66666666666666665}}\: \\ $$$${b}=\frac{\mathrm{999999999999999999999995}}{\mathrm{199999999999999999999999}}\: \\ $$$${In}\:{the}\:{way}\:{to}\:{cancel}\:,\:{if}\:{i}\:{just}\:{remove}\:{one}\:{common} \\ $$$${figure}\:{to}\:{the}\:{numerator}\:{and}\:{to}\:{the}\:{denominator} \\ $$$${And}\:{i}\:{find}\:{a}=\frac{\mathrm{2}}{\mathrm{5}}\:\:{and}\:\:{b}=\frac{\mathrm{5}}{\mathrm{1}}\: \\ $$$${Will}\:{it}\:{be}\:{wrong}?\:{if}\:\:{no}\:,{explain}! \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 65427    Answers: 0   Comments: 1

Question Number 61934    Answers: 1   Comments: 2

Answer: 0^0 =?

$$\mathrm{Answer}:\:\mathrm{0}^{\mathrm{0}} =? \\ $$

Question Number 60484    Answers: 1   Comments: 2

If a + b + c = 4 then find a^3 + b^3 + c^(3 ) = ?

$${If}\:\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{4} \\ $$$$ \\ $$$${then}\:{find} \\ $$$${a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}\:} =\:? \\ $$

Question Number 55057    Answers: 0   Comments: 0

Prove that Ln(z+1)=z−(z^2 /2)+(z^3 /3)−(z^4 /4)+... for ∣z∣< 1

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{Ln}\left({z}+\mathrm{1}\right)={z}−\frac{{z}^{\mathrm{2}} }{\mathrm{2}}+\frac{{z}^{\mathrm{3}} }{\mathrm{3}}−\frac{{z}^{\mathrm{4}} }{\mathrm{4}}+... \\ $$$$\mathrm{for}\:\mid{z}\mid<\:\mathrm{1} \\ $$

Question Number 47725    Answers: 1   Comments: 2

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