Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 51227 by mr W last updated on 25/Dec/18

How many odd numbers with different  digits are there from 2019 to 9102?

$${How}\:{many}\:{odd}\:{numbers}\:{with}\:{different} \\ $$$${digits}\:{are}\:{there}\:{from}\:\mathrm{2019}\:{to}\:\mathrm{9102}? \\ $$

Commented by mr W last updated on 25/Dec/18

1817?

$$\mathrm{1817}? \\ $$

Commented by Rasheed.Sindhi last updated on 26/Dec/18

You are very right sir!   See my answer below.(The  incorrect answer has been  deleted)

$${You}\:{are}\:{very}\:{right}\:{sir}!\: \\ $$$${See}\:{my}\:{answer}\:{below}.\left({The}\right. \\ $$$${incorrect}\:{answer}\:{has}\:{been} \\ $$$$\left.{deleted}\right) \\ $$

Commented by mr W last updated on 26/Dec/18

thanks alot sir for confirming and  for the hard work!

$${thanks}\:{alot}\:{sir}\:{for}\:{confirming}\:{and} \\ $$$${for}\:{the}\:{hard}\:{work}! \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/18

Sir how did you reech the answer?  Please share your approach.Your  approach might be more efficient!

$${Sir}\:{how}\:{did}\:{you}\:{reech}\:{the}\:{answer}? \\ $$$${Please}\:{share}\:{your}\:{approach}.{Your} \\ $$$${approach}\:{might}\:{be}\:{more}\:{efficient}! \\ $$

Answered by Rasheed.Sindhi last updated on 26/Dec/18

Case: Th=9  The biggest number is 9087  The smallest number 9013  th=9 (1 way)  h=0(1 way)  u={1,3,5,7,9}−{0,9}={1,3,5,7}(4 ways)  t={0,1,2,3,4,5,6,7,8,9}−{0,9,u}(7 ways)  ∴ Total ways(numbers) with 9 thousand  1×1×7×4=28  Case: Th=3,5,7  th={3,5,7}(3 ways)  u={1,3,5,7,9}−{th}(4 ways)  h={0,1,2,3,4,5,6,7,8,9}−{th,u}(8 ways)  t={0,1,2,3,4,5,6,7,8,9}−{th,u,h}(7 ways)  ∴ Total ways(numbers)          3×4×8×7=672  Case:Th=4,6,8  th={4,6,8}(3 ways)  u={1,3,5,7,9}−{th}={1,3,5,7,9}(5 ways)  h={0,1,2,3,4,5,6,7,8,9}−{th,u}(8 ways)  t={0,1,2,3,4,5,6,7,8,9}−{th,u,h}(7 ways)  ∴ Total ways(numbers)       3×5×8×7=840  Case: Th=2  2013 to 2987  th={2}(1 way)  u={1,3,5,7,9}(5ways)  h={0,1,2,3,4,5,6,7,8,9}−{2,u}(8 ways)  t={0,1,2,3,4,5,6,7,8,9}−{2,u,h}(7 ways)  ∴ Total ways(numbers)        1×5×8×7=280  Case: Th=2 ( up to 2019(exclusive) )  th={2}(1 way), h={0}(1 way)  t={0,1}−{0}={1}(1 way)  u={1,3,5,7,9}−{0,1,9}={3,5,7}(3 ways  Total ways(numbers) upto 2019 (except 2019)       1×1×1×3=3(to be subtracted)  [ These numbers are 2013,2015 & 2017 ]  Total of required numbers from 2019  to  9102        =28+672+840+280−3=1817

$${Case}:\:{Th}=\mathrm{9} \\ $$$${The}\:{biggest}\:{number}\:{is}\:\mathrm{9087} \\ $$$${The}\:{smallest}\:{number}\:\mathrm{9013} \\ $$$${th}=\mathrm{9}\:\left(\mathrm{1}\:{way}\right) \\ $$$${h}=\mathrm{0}\left(\mathrm{1}\:{way}\right) \\ $$$${u}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\}−\left\{\mathrm{0},\mathrm{9}\right\}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7}\right\}\left(\mathrm{4}\:{ways}\right) \\ $$$${t}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{\mathrm{0},\mathrm{9},{u}\right\}\left(\mathrm{7}\:{ways}\right) \\ $$$$\therefore\:{Total}\:{ways}\left({numbers}\right)\:{with}\:\mathrm{9}\:{thousand} \\ $$$$\mathrm{1}×\mathrm{1}×\mathrm{7}×\mathrm{4}=\mathrm{28} \\ $$$${Case}:\:{Th}=\mathrm{3},\mathrm{5},\mathrm{7} \\ $$$${th}=\left\{\mathrm{3},\mathrm{5},\mathrm{7}\right\}\left(\mathrm{3}\:{ways}\right) \\ $$$${u}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\}−\left\{{th}\right\}\left(\mathrm{4}\:{ways}\right) \\ $$$${h}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{{th},{u}\right\}\left(\mathrm{8}\:{ways}\right) \\ $$$${t}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{{th},{u},{h}\right\}\left(\mathrm{7}\:{ways}\right) \\ $$$$\therefore\:{Total}\:{ways}\left({numbers}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}×\mathrm{4}×\mathrm{8}×\mathrm{7}=\mathrm{672} \\ $$$${Case}:{Th}=\mathrm{4},\mathrm{6},\mathrm{8} \\ $$$${th}=\left\{\mathrm{4},\mathrm{6},\mathrm{8}\right\}\left(\mathrm{3}\:{ways}\right) \\ $$$${u}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\}−\left\{{th}\right\}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\}\left(\mathrm{5}\:{ways}\right) \\ $$$${h}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{{th},{u}\right\}\left(\mathrm{8}\:{ways}\right) \\ $$$${t}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{{th},{u},{h}\right\}\left(\mathrm{7}\:{ways}\right) \\ $$$$\therefore\:{Total}\:{ways}\left({numbers}\right) \\ $$$$\:\:\:\:\:\mathrm{3}×\mathrm{5}×\mathrm{8}×\mathrm{7}=\mathrm{840} \\ $$$${Case}:\:{Th}=\mathrm{2} \\ $$$$\mathrm{2013}\:{to}\:\mathrm{2987} \\ $$$${th}=\left\{\mathrm{2}\right\}\left(\mathrm{1}\:{way}\right) \\ $$$${u}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\}\left(\mathrm{5}{ways}\right) \\ $$$${h}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{\mathrm{2},{u}\right\}\left(\mathrm{8}\:{ways}\right) \\ $$$${t}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9}\right\}−\left\{\mathrm{2},{u},{h}\right\}\left(\mathrm{7}\:{ways}\right) \\ $$$$\therefore\:{Total}\:{ways}\left({numbers}\right) \\ $$$$\:\:\:\:\:\:\mathrm{1}×\mathrm{5}×\mathrm{8}×\mathrm{7}=\mathrm{280} \\ $$$${Case}:\:{Th}=\mathrm{2}\:\left(\:{up}\:{to}\:\mathrm{2019}\left({exclusive}\right)\:\right) \\ $$$${th}=\left\{\mathrm{2}\right\}\left(\mathrm{1}\:{way}\right),\:{h}=\left\{\mathrm{0}\right\}\left(\mathrm{1}\:{way}\right) \\ $$$${t}=\left\{\mathrm{0},\mathrm{1}\right\}−\left\{\mathrm{0}\right\}=\left\{\mathrm{1}\right\}\left(\mathrm{1}\:{way}\right) \\ $$$${u}=\left\{\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9}\right\}−\left\{\mathrm{0},\mathrm{1},\mathrm{9}\right\}=\left\{\mathrm{3},\mathrm{5},\mathrm{7}\right\}\left(\mathrm{3}\:{ways}\right. \\ $$$${Total}\:{ways}\left({numbers}\right)\:{upto}\:\mathrm{2019}\:\left({except}\:\mathrm{2019}\right) \\ $$$$\:\:\:\:\:\mathrm{1}×\mathrm{1}×\mathrm{1}×\mathrm{3}=\mathrm{3}\left({to}\:{be}\:{subtracted}\right) \\ $$$$\left[\:{These}\:{numbers}\:{are}\:\mathrm{2013},\mathrm{2015}\:\&\:\mathrm{2017}\:\right] \\ $$$${Total}\:{of}\:{required}\:{numbers}\:{from}\:\mathrm{2019} \\ $$$${to}\:\:\mathrm{9102}\: \\ $$$$\:\:\:\:\:=\mathrm{28}+\mathrm{672}+\mathrm{840}+\mathrm{280}−\mathrm{3}=\mathrm{1817} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com