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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-

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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.
$$\mathrm{The}\:\mathrm{conditions}\:\mathrm{are} \\ $$$$\frac{\mathrm{DEF}}{\mathrm{ABC}}\:=\:\mathrm{k}\:\left(\boldsymbol{\mathrm{integer}}\right)\:\: \\ $$$$\therefore\:\boldsymbol{\mathrm{k}}\left(\boldsymbol{\mathrm{n}}×\boldsymbol{\mathrm{ABC}}−\mathrm{1}\right)\:=\:\mathrm{1000}\:\left(\mathrm{n}=\:\mathrm{1},\mathrm{2},\mathrm{3},……\mathrm{9}\right) \\ $$$$\:\boldsymbol{\mathrm{k}}×\boldsymbol{\mathrm{ABC}}\:<\mathrm{1000} \\ $$$$\mathrm{then}…. \\ $$$$\mathrm{now}\:\mathrm{k}=\:\mathrm{2},\mathrm{4}\:\mathrm{and}\:\mathrm{8}\:\left(\mathrm{any}\:\mathrm{higher}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{1000}\:\mathrm{can}\:\mathrm{be}\:\right. \\ $$$$\left.\mathrm{cancelled}\right) \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{k}=\mathrm{2} \\ $$$$\mathrm{n}×\mathrm{ABC}=\mathrm{501}=\:\mathrm{3}×\mathrm{167} \\ $$$$\because\mathrm{167}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{the}\:\mathrm{only}\:\mathrm{possibility}\:\mathrm{is} \\ $$$$\mathrm{2}\left(\mathrm{3×167}−\mathrm{1}\right)\:\mathrm{i}.\mathrm{e}.\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{167}\left(\mathrm{2×167}\right) \\ $$$$=\mathrm{167334}\:=\:\mathrm{3}×\mathrm{167}×\mathrm{334} \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{k}=\mathrm{4},\:\:\boldsymbol{\mathrm{n}}×\boldsymbol{\mathrm{ABC}}=\mathrm{251}\:\left(\mathrm{which}\:\mathrm{is}\:\mathrm{prime}\right) \\ $$$$\because\:\mathrm{4}×\mathrm{251}>\mathrm{1000}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{such}\:\mathrm{number}\:\mathrm{exists}. \\ $$$$\mathrm{for}\:\mathrm{k}=\:\mathrm{8}\:,\:\boldsymbol{\mathrm{n}}×\boldsymbol{\mathrm{ABC}}=\:\mathrm{126}\left(\mathrm{2}×\mathrm{3}×\mathrm{3}×\mathrm{7}\right) \\ $$$$\therefore\mathrm{the}\:\mathrm{numbers}\:\mathrm{are} \\ $$$$\mathrm{8}\left(\mathrm{2×63}−\mathrm{1}\right)\Rightarrow\:\mathrm{063504}\:=\:\mathrm{2}×\mathrm{063}×\mathrm{504} \\ $$$$\mathrm{8}\left(\mathrm{3}×\mathrm{42}−\mathrm{1}\right)\Rightarrow\mathrm{042336}\:=\:\mathrm{3}×\mathrm{042}×\mathrm{336} \\ $$$$\mathrm{8}\left(\mathrm{7}×\mathrm{18}−\mathrm{1}\right)\Rightarrow\mathrm{018144}=\:\mathrm{7}×\mathrm{018}×\mathrm{144} \\ $$$$\mathrm{8}\left(\mathrm{9}×\mathrm{14}−\mathrm{1}\right)\Rightarrow\mathrm{014112}\:=\:\mathrm{9}×\mathrm{014}×\mathrm{112} \\ $$$$\because\:\mathrm{the}\:\mathrm{digits}\:\mathrm{are}\:\:\mathrm{all}\:\mathrm{repeated}\:. \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
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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