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so a 4 digit number is represented ABCD, they are not multiplied together, each letter represents 1 single , whole number

if ABCD times D=DCBA

solve for the values of A and B and C and D with full workings and solution (this problem is from my 5th grade math book, just for fun)

Sagot :

[tex]\overline{ABCD}=10^3A+10^2B+10C+D\\\\\overline{ABCD}\times D=(10^3A+10^2B+10C+D)\times D\\=10^3AD+10^2BD+10CD+D^2\\\\\overline{DCBA}=10^3D+10^2C+10B+A\\\\\overline{ABCD}\times D=\overline{DCBA}\\\downarrow\\10^3AD+10^2BD+10CD+D^2=10^3D+10^2C+10B+A[/tex]

[tex]one\ of\ the\ solutions\ of\ general:\\\\\#1\ AD=D\\\#2\ BD=C\\\#3\ CD=B\\\#4\ D^2=A\\\\from\ \#1\ and\ \#4\\AD=D\ and\ D^2=A\to\ conclusion:A=D=1\\\\from\ \#2\ and\ \#3\\BD=C\ and\ CD=B\to conclusion:B=C\\\\therefore[/tex]

[tex]\overline{ABCD}=\\1001\\1111\\1221\\1331\\1441\\1551\\1661\\1771\\1881\\1991[/tex]

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