\(101=50+51\) (som van opeenvolgende gehele getallen)

\(101=13+17+19+23+29\) (som van opeenvolgende priemgetallen)

\(101=5!-4!+3!-2!+1!\)

\(101=((0;0;1;10)\,(0;1;6;8)\,(0;2;4;9)\,(0;4;6;7)\,(2;5;6;6))\lower2pt{\Large{\color{teal}{➋}}}\to\{\#5\}\)

\(101\mathbf{\color{blue}{\;=\;}}10^2+1\)

\(101=((0;0;0;0;1;1;2;3;4)\,(1;1;1;1;2;2;3;3;3))\lower2pt{\Large{\color{teal}{➌}}}\to\{\#2\}\)

\(101=3^3+5^2+7^2\)

\(101=\Large\frac{\sqrt{101^3\,-\,101^2}}{\lfloor\sqrt{101}\,\rfloor}\)

\(101\mathbf{\color{blue}{\;=\;}}2^7-3^3\mathbf{\color{blue}{\;=\;}}\bbox[2px,border:1px brown dashed]{51^2-50^2}\mathbf{\color{blue}{\;=\;}}926^2-95^3\)

101.1

\(101\mathbf{\color{blue}{\;=\;}}\)(som van drie derdemachten)

\(\qquad~~~~10\) oplossingen bekend

\(\qquad~~~~\)References Sum of Three Cubes

\(\qquad~~~~\bbox[3px,border:1px solid]{(-3)^3+4^3+4^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{46^3+78^3+(-83)^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{(-251)^3+(-272)^3+330^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{(-167)^3+(-621)^3+625^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{319^3+2325^3+(-2327)^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{(-21923)^3+(-39318)^3+41470^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{(-259856813)^3+(-950791085)^3+957217647^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{859962468^3+1066448860^3+(-1227348611)^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{14539096087636^3+106817695040893^3+(-106907404837908)^3}\mathbf{\color{blue}{\;=\;}}\)

\(\qquad~~~~\bbox[3px,border:1px solid]{44903816710797^3+(-139549058795306)^3+137981724756814^3}\mathbf{\color{blue}{\;=\;}}\)

\(101\mathbf{\color{blue}{\;=\;}}\)(som van vijf vijfdemachten)

\(\qquad~~~~(z\gt200)\)

\(\qquad~~~~\bbox[3px,border:1px blue solid]{(-55)^5+(-226)^5+461^5+511^5+(-560)^5}\)

\(\qquad~~~~\bbox[3px,border:1px blue solid]{411^5+(-475)^5+(-550)^5+(-576)^5+661^5}\)

101.2

 ○–○–○ 

\(101^2=10201~~\) en \(~~102+0-1=101\)

\(101^3=1030301~~\) en \(~~103+0-3+0+1=101\)

\(101^4=104060401~~\) en \(~~104+0-6+0+4+0-1=101\)

\(101^5=10510100501~~\) en \(~~105-10+10+0-5+0+1=101\)

\(101^6=1061520150601~~\) en \(~~106-15+20-15+0+6+0-1=101\)

\(101^7=107213535210701~~\) en \(~~107-21+3+5+3-5+2-1+0+7+0+1=101\)

\(101^8=10828567056280801~~\) en \(~~1+0-8-2+8+5+67+0+5+6+2+8+0+8+0+1=101\)

\(101^9=1093685272684360901~~\) en \(~~109-36+85-27+26-84+36+0-9+01=101\)

101.3

\begin{align} 101^2&=1~02~01\\ 101^3&=1~03~03~01\\ 101^4&=1~04~06~04~01\\ 101^5&=1~05~10~10~05~01\\ 101^6&=1~06~15~20~15~06~01\\ 101^7&=1~07~21~35~35~21~07~01\\ 101^8&=1~08~28~56~70~56~28~08~01\\ Vanaf~101^9&=1.093.685.272.684.360.901~wordt~het~patroon~doorbroken \end{align}

Tot \(101^8\) krijgt men de coëfficiënten van de driehoek van PASCAL.
Zie hoofdstuk Driehoek van PASCAL uit “Getallen in Detail”. 		101.4
  EEN WEETJE  

\(101\) en \(110\) hebben dezelfde cijfers. Zo ook \(101^2 = 10201\) en \(110^2 = 12100\) evenals de derdemacht : \(101^3 = 1030301\)
en \(110^3 = 1331000\). Tenslotte heeft men ook nog \(101^4 = 104060401\) en \(110^4 = 14641000\).
Merk op dat \(101^n\) een palindroom is en \(110^n\) met weglating van de eindnullen eveneens palindroom is.
Voor hogere machten lukt dit niet meer (d.w.z. van zodra de coëfficiënten in de driehoek van Pascal groter dan \(09\) zijn
en dus met een cijfer \(1\) of hoger beginnen). Ter verificatie volgen hier enkele waarden :

\(101^5=10510100501~~\) en \(~~110^5=16105100000\)

\(101^6=10615201500601~~\) en \(~~110^6=1771561000000\)

\(101^7=107213535210701~~\) en \(~~110^7=194871710000000\)

\(101^8=10828568056280801~~\) en \(~~110^8=21435888100000000\)

\(101^9=1093685272684360901~~\) en \(~~110^9=2357947691000000000\)

101.5
Men moet \(101\) tot minimaal de \(36208\)ste macht verheffen opdat in de decimale expansie exact \(101\) \(101\)'s verschijnen.
Terloops : \(101\)\(^{36208}\) heeft een lengte van \(72573\) cijfers.
Om aan het juiste aantal van \(101\) te komen moet je hier wel rekening houden met overlappingen. Zo hebben we
\(96\) maal \(101\) (incl. \(101|{\color{grey}{01}}\)) en \(5\) maal \({\color{grey}{10}}|101\) wat ons totaal op \(101\) brengt (dit fenomeen doet zich enkel voor met
repdigits en palindromen).
Zonder overlappingen moeten we de exponent laten oplopen tot \(37079\). En \(101\)\(^{37079}\) is dan \(74319\) cijfers lang.
De grootste exponent die nog een oplossing zonder overlappingen geeft is \(65850\). Hogerop wacht slechts de \(\large{\infty}\).
101.6

\(101^2\mathbf{\color{blue}{\;=\;}}20^2+99^2\mathbf{\color{blue}{\;\mathbf{\color{blue}{\;=\;}}\;}}{\color{blue}{101}}^3-1010^2\mathbf{\color{blue}{\;\mathbf{\color{blue}{\;=\;}}\;}}155^2-24^3\mathbf{\color{blue}{\;\mathbf{\color{blue}{\;=\;}}\;}}5101^2-5100^2\)

\(101^3\mathbf{\color{blue}{\;=\;}}{\color{blue}{101}}^2+1010^2\mathbf{\color{blue}{\;=\;}}299^2+970^2\mathbf{\color{blue}{\;=\;}}\bbox[2px,border:1px brown dashed]{5151^2-5050^2}\)

101.7

\begin{align} 101^2&=102\mathbf{01}\\ 201^2&=404\mathbf{01}\\ 301^2&=906\mathbf{01}\\ 401^2&=1608\mathbf{01}\\ 501^2&=2510\mathbf{01}\\ \cdots&=\cdots \end{align}

Het patroon is steeds hetzelfde : een kwadraat, een even getal en als eindcijfers \(01\) 		101.8

De eerste keer dat er \(101\) opeenvolgende samengestelde getallen voorkomen gebeurt tussen de priemgetallen \(1444309\)
en \(1444411\) met aldus een priemkloof van \(102\,.~~\) (OEIS A000101.pdf)

101.9
De palindroomgetallen aaneengeschakeld vanaf \(1\) tot en met \(101\) vormen een priemgetal. Dit is de tweede maal dat
dit gebeurt. Zie bij . Het priemgetal is \(123456789112233445566778899{\color{blue}{101}}\).
Het derde getal in de reeks is nog onbekend. 		101.10

\(3\)\(^{101}\)\(~=~1546132562196033993109383389296863818106322566003\) is de hoogst gekende macht van \(3\) waarbij

geen cijfer \(7\) voorkomt in de decimale expansie. (OEIS A131615)

101.11

\(101\) is de som van alle producten \(p*q\) waar alleen de vier eerste priemgetallen bij betrokken zijn.

\(101=2*3+2*5+2*7+3*5+3*7+5*7\)

101.12

Som der reciproken van partitiegetallen van \(101\) is \(1\) op \(118\) (honderdachttien) wijzen.

Vijf partities hebben unieke termen.

\(~~(1)~~\bbox[navajowhite,3px,border:1px solid]{101=2+4+5+30+60}~~\) en \(~~{1=\Large\frac{1}{2}}+{\Large\frac{1}{4}}+{\Large\frac{1}{5}}+{\Large\frac{1}{30}}+{\Large\frac{1}{60}}\)

\(~~(7)~~\bbox[navajowhite,3px,border:1px solid]{101=2+5+10+15+20+21+28}~~\) en \(~~{1=\Large\frac{1}{2}}+{\Large\frac{1}{5}}+{\Large\frac{1}{10}}+{\Large\frac{1}{15}}+{\Large\frac{1}{20}}+{\Large\frac{1}{21}}+{\Large\frac{1}{28}}\)

\((11)~~\bbox[navajowhite,3px,border:1px solid]{101=2+6+8+12+21+24+28}~~\) en \(~~{1=\Large\frac{1}{2}}+{\Large\frac{1}{6}}+{\Large\frac{1}{8}}+{\Large\frac{1}{12}}+{\Large\frac{1}{21}}+{\Large\frac{1}{24}}+{\Large\frac{1}{28}}\)

\((12)~~\bbox[navajowhite,3px,border:1px solid]{101=2+6+9+11+18+22+33}~~\) en \(~~{1=\Large\frac{1}{2}}+{\Large\frac{1}{6}}+{\Large\frac{1}{9}}+{\Large\frac{1}{11}}+{\Large\frac{1}{18}}+{\Large\frac{1}{22}}+{\Large\frac{1}{33}}\)

\((19)~~\bbox[navajowhite,3px,border:1px solid]{101=3+4+5+10+21+28+30}~~\) en \(~~{1=\Large\frac{1}{3}}+{\Large\frac{1}{4}}+{\Large\frac{1}{5}}+{\Large\frac{1}{10}}+{\Large\frac{1}{21}}+{\Large\frac{1}{28}}+{\Large\frac{1}{30}}\)

(OEIS A125726)

101.13

\(b\mathbf{\color{blue}{\;=\;}}101\to b\)\(^{2}\)\(+b\)\(^{2}\)\(+b\)\(^{0}\)\(+b\)\(^{9}\)\(+b\)\(^{2}\)\(+b\)\(^{4}\)\(+b\)\(^{3}\)\(+b\)\(^{1}\)\(+b\)\(^{6}\)\(+b\)\(^{8}\)\(+b\)\(^{1}\)\(+b\)\(^{7}\)\(+b\)\(^{8}\)\(+b\)\(^{0}\)\(+b\)\(^{0}\)\(+b\)\(^{7}\)\(+b\)\(^{3}\)\(+b\)\(^{1}\)\(+b\)\(^{9}\)\(\mathbf{\color{blue}{\;=\;}}2209243168178007319~~\)
(OEIS A236067)

101.14

\(101\) als expressie met de cijfers van

\(1\) tot \(9\) in stijgende volgorde

\(11\) oplossingen

\(101=12+34-56-7+8+9\)

\(101=1+2+3+45+67-8-9\)

\(101=1+2+34+56+7-8+9\)

\(101=1+2-3+4-5+6+7+89\)

\(101=1+23+4+5+67-8+9\)

\(101=1+23-4+5-6-7+89\)

\(101=1-2+3+4+5-6+7+89\)

\(101=1-2+34+5-6+78-9\)

\(101=12+34+5+67-8-9\)

\(101=12-3+4-5+6+78+9\)

\(101=123+4+56+7-89\)

\(101=1+2+34+5+6*7+8+9\)

\(101\) als expressie met de cijfers van

\(1\) tot \(9\) in willeurige volgorde

\(79\) oplossingen

\(101=(2+6)*9+(8-3)/5+4^1*7\)

\(101=(2+6)*9+(8-3)/5+7^1*4\)

\(101=(2+6)*9+(8-5)/3+4^1*7\)

\(101=(2+6)*9+(8-5)/3+7^1*4\)

\(101=(3+5)*9+(8-2)/6+4^1*7\)

\(101=(3+5)*9+(8-2)/6+7^1*4\)

\(101=(3+5)*9+(8-6)/2+4^1*7\)

\(101=(3+5)*9+(8-6)/2+7^1*4\)

\(101=(3+8)*6+(2-7)/5+4^1*9\)

\(101=(3+8)*6+(2-7)/5+9^1*4\)

\(101=(3+8)*6+(5-7)/2+4^1*9\)

\(101=(3+8)*6+(5-7)/2+9^1*4\)

\(101=(3+9)*8+(2-7)/5+1^4*6\)

\(101=(3+9)*8+(5-7)/2+1^4*6\)

\(101=(3+9)*8+(7-2)/5+1^6*4\)

\(101=(3+9)*8+(7-5)/2+1^6*4\)

\(101=(4+5)*9+(2-8)/6+3^1*7\)

\(101=(4+5)*9+(2-8)/6+7^1*3\)

\(101=(4+5)*9+(6-8)/2+3^1*7\)

\(101=(4+5)*9+(6-8)/2+7^1*3\)

\(101=(4+6)*3+(2-7)/5+8^1*9\)

\(101=(4+6)*3+(2-7)/5+9^1*8\)

\(101=(4+6)*3+(5-7)/2+8^1*9\)

\(101=(4+6)*3+(5-7)/2+9^1*8\)

\(101=(4+7)*9+(2-8)/6+1^5*3\)

\(101=(4+7)*9+(6-8)/2+1^5*3\)

\(101=(4+9)*6+(2-7)/5+3^1*8\)

\(101=(4+9)*6+(2-7)/5+8^1*3\)

\(101=(4+9)*6+(5-7)/2+3^1*8\)

\(101=(4+9)*6+(5-7)/2+8^1*3\)

\(101=(5+6)*4+(9-1)/8+2^3*7\)

\(101=(5+6)*4+(9-8)/1+2^3*7\)

\(101=(5+6)*8+(9-2)/7+3^1*4\)

\(101=(5+6)*8+(9-2)/7+4^1*3\)

\(101=(5+6)*8+(9-7)/2+3^1*4\)

\(101=(5+6)*8+(9-7)/2+4^1*3\)

\(101=(5+6)*9+(2-8)/3+1^7*4\)

\(101=(5+6)*9+(3-7)/2+1^8*4\)

\(101=(5+7)*4+(1-9)/8+3^2*6\)

\(101=(5+7)*4+(8-9)/1+3^2*6\)

\(101=(5+7)*6+(3-9)/2+4^1*8\)

\(101=(5+7)*6+(3-9)/2+8^1*4\)

\(101=(5+7)*6+(8-4)/2+3^1*9\)

\(101=(5+7)*6+(8-4)/2+9^1*3\)

\(101=(5+7)*8+(9-3)/6+1^2*4\)

\(101=(5+7)*8+(9-6)/3+1^2*4\)

\(101=(5+8)*3+(2-6)/4+7^1*9\)

\(101=(5+8)*3+(2-6)/4+9^1*7\)

\(101=(5+8)*3+(4-6)/2+7^1*9\)

\(101=(5+8)*3+(4-6)/2+9^1*7\)

\(101=(5+8)*7+(6-2)/4+1^3*9\)

\(101=(5+8)*7+(6-4)/2+1^3*9\)

\(101=(5+9)*6+(7-3)/4+2^1*8\)

\(101=(5+9)*6+(7-3)/4+8^1*2\)

\(101=(5+9)*6+(7-4)/3+2^1*8\)

\(101=(5+9)*6+(7-4)/3+8^1*2\)

\(101=(5+9)*7+(2-8)/6+1^3*4\)

\(101=(5+9)*7+(6-8)/2+1^3*4\)

\(101=(6+8)*5+(9-3)/2+4^1*7\)

\(101=(6+8)*5+(9-3)/2+7^1*4\)

\(101=(6+8)*7+(2-5)/3+1^9*4\)

\(101=(6+8)*7+(3-5)/2+1^9*4\)

\(101=(6+8)*7+(9-4)/5+1^3*2\)

\(101=(6+8)*7+(9-5)/4+1^3*2\)

\(101=(6+9)*4+(8-1)/7+2^3*5\)

\(101=(6+9)*4+(8-7)/1+2^3*5\)

\(101=(6+9)*5+(2-8)/3+4^1*7\)

\(101=(6+9)*5+(2-8)/3+7^1*4\)

\(101=(7+8)*5+(2-6)/4+3^1*9\)

\(101=(7+8)*5+(2-6)/4+9^1*3\)

\(101=(7+8)*5+(4-6)/2+3^1*9\)

\(101=(7+8)*5+(4-6)/2+9^1*3\)

\(101=(7+9)*4+(8-3)/5+6^2*1\)

\(101=(7+9)*4+(8-5)/3+6^2*1\)

\(101=(7+9)*6+(5-2)/3+1^8*4\)

\(101=(7+9)*6+(5-3)/2+1^8*4\)

\(101=(7+9)*6+(8-3)/5+1^2*4\)

\(101=(7+9)*6+(8-4)/2+1^5*3\)

\(101=(7+9)*6+(8-5)/3+1^2*4\)

Met de cijfers van \(1\) tot \(9\) in stijgende en dalende volgorde (met dank aan Inder. J. Taneja) :
\(\qquad\qquad~~101=1+2+34+5+6*7+8+9\)
\(\qquad\qquad~~101=9*8+7+6+5+4+3*2+1\)

101.15

Som Der Cijfers (\(sdc\)) van \(k^{\large{101}}\) is gelijk aan het grondtal \(k\). De triviale oplossingen \(0\) en \(1\) negerend vinden we :

\(\qquad\qquad~sdc\left(1423^{\large{101}}\right)=1423\qquad\qquad~sdc\left(1468^{\large{101}}\right)=1468\)

101.16

Expressie met tweemaal de cijfers uit het getal \(101\)
\(101=(1\)^^\(0\)^^\(1)+1+0-1\)

101.17

Als expressie met enkelcijferige toepassing, resp. van \(1\) tot \(9~~\) (met dank aan Inder. J. Taneja).
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}1111/11\mathbf{\color{blue}{\;=\;}}(11-1)^{(1+1)}+1\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}2222/22\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}3333/33\mathbf{\color{blue}{\;=\;}}3+3*33-3/3\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}4444/44\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}5555/55\mathbf{\color{blue}{\;=\;}}5*(5*5-5)+5/5\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}6666/66\mathbf{\color{blue}{\;=\;}}66+6*6-6/6\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}7777/77\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}8888/88\)
\(\qquad\qquad101\mathbf{\color{blue}{\;=\;}}9999/99\mathbf{\color{blue}{\;=\;}}99+(9+9)/9\)

101.18
\(101\) is het aantal partities van \(13~~\) (OEIS A000041)
Pari/GP code : numbpart(13) 		101.19
\(101\) is de som van de eerste \(8\) semipriemgetallen \((4+6+9+10+14+15+21+22)\).
(OEIS A062198) 		101.20
Het kleinste getal dat exact \(101\) delers heeft is \(1267650600228229401496703205376=2^{100}\). (OEIS A005179) 101.21
Schakelaar
\(\mathbf[0\gets\to1000\mathbf]\)
“Allemaal Getallen”

\(\Huge\bbox[border:0]{⏮}\)

\(\Huge\bbox[border:0]{⯬}\)

\(\Huge\bbox[border:0]{⏴}\)

\(\Huge\bbox[border:0]{⏵}\)

\(\Huge\bbox[border:0]{⯮}\)

\(\Huge\bbox[border:0]{⏭}\)

\(101\)\(_{\large\color{green}{26}}\)\(101\)\(2\)\(102\)
\(1,101\)
Priemgetal\(1100101_2\)\(65_{16}\)
   

Uit de collectie 'Allemaal Getallen' van Ir. Jos Heynderickx
Toevoegingen & Bewerking & Layout door Patrick De Geest (email)
Laatste update 24 april 2025