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Question Number 91390 by mr W last updated on 30/Apr/20

ABCDEF is a 6 digit number,  ABC and DEF are 3 digit numbers.  find ABCDEF  satisfying:  1)   ABCDEF=1×ABC×DEF  2)   ABCDEF=2×ABC×DEF  3)   ABCDEF=3×ABC×DEF  4)   ABCDEF=4×ABC×DEF  5)   ABCDEF=5×ABC×DEF  6)   ABCDEF=6×ABC×DEF  7)   ABCDEF=7×ABC×DEF  8)   ABCDEF=8×ABC×DEF  9)   ABCDEF=9×ABC×DEF

$${ABCDEF}\:{is}\:{a}\:\mathrm{6}\:{digit}\:{number}, \\ $$$${ABC}\:{and}\:{DEF}\:{are}\:\mathrm{3}\:{digit}\:{numbers}. \\ $$$${find}\:{ABCDEF}\:\:{satisfying}: \\ $$$$\left.\mathrm{1}\right)\:\:\:{ABCDEF}=\mathrm{1}×{ABC}×{DEF} \\ $$$$\left.\mathrm{2}\right)\:\:\:{ABCDEF}=\mathrm{2}×{ABC}×{DEF} \\ $$$$\left.\mathrm{3}\right)\:\:\:{ABCDEF}=\mathrm{3}×{ABC}×{DEF} \\ $$$$\left.\mathrm{4}\right)\:\:\:{ABCDEF}=\mathrm{4}×{ABC}×{DEF} \\ $$$$\left.\mathrm{5}\right)\:\:\:{ABCDEF}=\mathrm{5}×{ABC}×{DEF} \\ $$$$\left.\mathrm{6}\right)\:\:\:{ABCDEF}=\mathrm{6}×{ABC}×{DEF} \\ $$$$\left.\mathrm{7}\right)\:\:\:{ABCDEF}=\mathrm{7}×{ABC}×{DEF} \\ $$$$\left.\mathrm{8}\right)\:\:\:{ABCDEF}=\mathrm{8}×{ABC}×{DEF} \\ $$$$\left.\mathrm{9}\right)\:\:\:{ABCDEF}=\mathrm{9}×{ABC}×{DEF} \\ $$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

Yes Sir for k=1  Actually I just not consider 1 as the divisor of  1000. Thank you sir.

$$\mathrm{Yes}\:\mathrm{Sir}\:\mathrm{for}\:\mathrm{k}=\mathrm{1} \\ $$$$\mathrm{Actually}\:\mathrm{I}\:\mathrm{just}\:\mathrm{not}\:\mathrm{consider}\:\mathrm{1}\:\mathrm{as}\:\mathrm{the}\:\mathrm{divisor}\:\mathrm{of} \\ $$$$\mathrm{1000}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

The conditions are  ((DEF)/(ABC)) = k (integer)    ∴ k(n×ABC−1) = 1000 (n= 1,2,3,......9)   k×ABC <1000  then....  now k= 2,4 and 8 (any higher multiple of 1000 can be   cancelled)  now for k=2  n×ABC=501= 3×167  ∵167 is a prime the only possibility is  2(3x167−1) i.e.the number is 167(2x167)  =167334 = 3×167×334  now for k=4,  n×ABC=251 (which is prime)  ∵ 4×251>1000 there is no such number exists.  for k= 8 , n×ABC= 126(2×3×3×7)  ∴the numbers are  8(2x63−1)⇒ 063504 = 2×063×504  8(3×42−1)⇒042336 = 3×042×336  8(7×18−1)⇒018144= 7×018×144  8(9×14−1)⇒014112 = 9×014×112  ∵ the digits are  all repeated .  please check.

$$\mathrm{The}\:\mathrm{conditions}\:\mathrm{are} \\ $$$$\frac{\mathrm{DEF}}{\mathrm{ABC}}\:=\:\mathrm{k}\:\left(\boldsymbol{\mathrm{integer}}\right)\:\: \\ $$$$\therefore\:\boldsymbol{\mathrm{k}}\left(\boldsymbol{\mathrm{n}}×\boldsymbol{\mathrm{ABC}}−\mathrm{1}\right)\:=\:\mathrm{1000}\:\left(\mathrm{n}=\:\mathrm{1},\mathrm{2},\mathrm{3},......\mathrm{9}\right) \\ $$$$\:\boldsymbol{\mathrm{k}}×\boldsymbol{\mathrm{ABC}}\:<\mathrm{1000} \\ $$$$\mathrm{then}.... \\ $$$$\mathrm{now}\:\mathrm{k}=\:\mathrm{2},\mathrm{4}\:\mathrm{and}\:\mathrm{8}\:\left(\mathrm{any}\:\mathrm{higher}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{1000}\:\mathrm{can}\:\mathrm{be}\:\right. \\ $$$$\left.\mathrm{cancelled}\right) \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{k}=\mathrm{2} \\ $$$$\mathrm{n}×\mathrm{ABC}=\mathrm{501}=\:\mathrm{3}×\mathrm{167} \\ $$$$\because\mathrm{167}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{the}\:\mathrm{only}\:\mathrm{possibility}\:\mathrm{is} \\ $$$$\mathrm{2}\left(\mathrm{3x167}−\mathrm{1}\right)\:\mathrm{i}.\mathrm{e}.\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{167}\left(\mathrm{2x167}\right) \\ $$$$=\mathrm{167334}\:=\:\mathrm{3}×\mathrm{167}×\mathrm{334} \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{k}=\mathrm{4},\:\:\boldsymbol{\mathrm{n}}×\boldsymbol{\mathrm{ABC}}=\mathrm{251}\:\left(\mathrm{which}\:\mathrm{is}\:\mathrm{prime}\right) \\ $$$$\because\:\mathrm{4}×\mathrm{251}>\mathrm{1000}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{such}\:\mathrm{number}\:\mathrm{exists}. \\ $$$$\mathrm{for}\:\mathrm{k}=\:\mathrm{8}\:,\:\boldsymbol{\mathrm{n}}×\boldsymbol{\mathrm{ABC}}=\:\mathrm{126}\left(\mathrm{2}×\mathrm{3}×\mathrm{3}×\mathrm{7}\right) \\ $$$$\therefore\mathrm{the}\:\mathrm{numbers}\:\mathrm{are} \\ $$$$\mathrm{8}\left(\mathrm{2x63}−\mathrm{1}\right)\Rightarrow\:\mathrm{063504}\:=\:\mathrm{2}×\mathrm{063}×\mathrm{504} \\ $$$$\mathrm{8}\left(\mathrm{3}×\mathrm{42}−\mathrm{1}\right)\Rightarrow\mathrm{042336}\:=\:\mathrm{3}×\mathrm{042}×\mathrm{336} \\ $$$$\mathrm{8}\left(\mathrm{7}×\mathrm{18}−\mathrm{1}\right)\Rightarrow\mathrm{018144}=\:\mathrm{7}×\mathrm{018}×\mathrm{144} \\ $$$$\mathrm{8}\left(\mathrm{9}×\mathrm{14}−\mathrm{1}\right)\Rightarrow\mathrm{014112}\:=\:\mathrm{9}×\mathrm{014}×\mathrm{112} \\ $$$$\because\:\mathrm{the}\:\mathrm{digits}\:\mathrm{are}\:\:\mathrm{all}\:\mathrm{repeated}\:. \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$

Commented by mr W last updated on 30/Apr/20

an other one is  143143=7×143×143

$${an}\:{other}\:{one}\:{is} \\ $$$$\mathrm{143143}=\mathrm{7}×\mathrm{143}×\mathrm{143} \\ $$

Answered by mr W last updated on 30/Apr/20

say ABC=x≥100, DEF=y≥100  1000x+y=kyx with k=1,2,...,9  1000+(y/x)=ky  say y=nx, 1≤n≤9  1000=(kx−1)n  n=1:  kx=1001=7×11×13  k=1⇒x=1001 bad  k=7⇒x=143⇒y=143 good    n=2:  kx=501=3×167  k=1⇒x=501⇒y=1002 bad  k=3⇒x=167⇒y=334 good    n=4:  kx=251=prime  k=1⇒x=251⇒y=1004 bad    n=5:  kx=201=3×67  k=1⇒x=201⇒y=1005 bad    n=8:  kx=126=2×3^2 ×7  k=1⇒x=126⇒y=1008 bad    summary:  two solutions:  167334=3×167×334  143143=7×143×143

$${say}\:{ABC}={x}\geqslant\mathrm{100},\:{DEF}={y}\geqslant\mathrm{100} \\ $$$$\mathrm{1000}{x}+{y}={kyx}\:{with}\:{k}=\mathrm{1},\mathrm{2},...,\mathrm{9} \\ $$$$\mathrm{1000}+\frac{{y}}{{x}}={ky} \\ $$$${say}\:{y}={nx},\:\mathrm{1}\leqslant{n}\leqslant\mathrm{9} \\ $$$$\mathrm{1000}=\left({kx}−\mathrm{1}\right){n} \\ $$$${n}=\mathrm{1}: \\ $$$${kx}=\mathrm{1001}=\mathrm{7}×\mathrm{11}×\mathrm{13} \\ $$$${k}=\mathrm{1}\Rightarrow{x}=\mathrm{1001}\:{bad} \\ $$$${k}=\mathrm{7}\Rightarrow{x}=\mathrm{143}\Rightarrow{y}=\mathrm{143}\:{good} \\ $$$$ \\ $$$${n}=\mathrm{2}: \\ $$$${kx}=\mathrm{501}=\mathrm{3}×\mathrm{167} \\ $$$${k}=\mathrm{1}\Rightarrow{x}=\mathrm{501}\Rightarrow{y}=\mathrm{1002}\:{bad} \\ $$$${k}=\mathrm{3}\Rightarrow{x}=\mathrm{167}\Rightarrow{y}=\mathrm{334}\:{good} \\ $$$$ \\ $$$${n}=\mathrm{4}: \\ $$$${kx}=\mathrm{251}={prime} \\ $$$${k}=\mathrm{1}\Rightarrow{x}=\mathrm{251}\Rightarrow{y}=\mathrm{1004}\:{bad} \\ $$$$ \\ $$$${n}=\mathrm{5}: \\ $$$${kx}=\mathrm{201}=\mathrm{3}×\mathrm{67} \\ $$$${k}=\mathrm{1}\Rightarrow{x}=\mathrm{201}\Rightarrow{y}=\mathrm{1005}\:{bad} \\ $$$$ \\ $$$${n}=\mathrm{8}: \\ $$$${kx}=\mathrm{126}=\mathrm{2}×\mathrm{3}^{\mathrm{2}} ×\mathrm{7} \\ $$$${k}=\mathrm{1}\Rightarrow{x}=\mathrm{126}\Rightarrow{y}=\mathrm{1008}\:{bad} \\ $$$$ \\ $$$${summary}: \\ $$$${two}\:{solutions}: \\ $$$$\mathrm{167334}=\mathrm{3}×\mathrm{167}×\mathrm{334} \\ $$$$\mathrm{143143}=\mathrm{7}×\mathrm{143}×\mathrm{143} \\ $$

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