BdMO National Junior 2007/3
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
A man has $4$ childern.The age of the first child is a square number.By multiplying the digits of this square number you will get the age of second child and by summing up the digits you will get the age of third child.If you add the digits of the age of the second child,you will get the age of fourth child.If the difference of age of two consequetive childern is not more than $25$ years,then find the ages if $4$ children.
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2007/3
\[6^2=36\]
\[2×6=12\]
\[2+6=8\]
\[1+2=3\]
\[36>12>8>3 \]
Age of,
\[ first child =36\]
\[ second child =12\]
\[third child =8\]
\[fourth child =3\]
\[2×6=12\]
\[2+6=8\]
\[1+2=3\]
\[36>12>8>3 \]
Age of,
\[ first child =36\]
\[ second child =12\]
\[third child =8\]
\[fourth child =3\]