Difference between revisions of "2021 CIME I Problems/Problem 2"

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==Problem 2==
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== Problem ==
 
For digits <math>a, b, c,</math> with <math>a\neq 0,</math> the positive integer <math>N</math> can be written as <math>\underline{a}\underline{a}\underline{b}\underline{b}</math> in base <math>9,</math> and <math>\underline{a}\underline{a}\underline{b}\underline{b}\underline{c}</math> in base <math>5</math>. Find the base-<math>10</math> representation of <math>N</math>.
 
For digits <math>a, b, c,</math> with <math>a\neq 0,</math> the positive integer <math>N</math> can be written as <math>\underline{a}\underline{a}\underline{b}\underline{b}</math> in base <math>9,</math> and <math>\underline{a}\underline{a}\underline{b}\underline{b}\underline{c}</math> in base <math>5</math>. Find the base-<math>10</math> representation of <math>N</math>.
  

Latest revision as of 19:00, 24 July 2024

Problem

For digits $a, b, c,$ with $a\neq 0,$ the positive integer $N$ can be written as $\underline{a}\underline{a}\underline{b}\underline{b}$ in base $9,$ and $\underline{a}\underline{a}\underline{b}\underline{b}\underline{c}$ in base $5$. Find the base-$10$ representation of $N$.

Solution

Consider the different representations of the number and equate them: \[(9^3 + 9^2) a +(9+1)b = (5^4+5^3) a + (5^2+5)b+c\] \[(810)a+10b = (750)a+30b+c\] \[60 a - 20b-c=0\]

Note that c can't contribute since it is less than 5 so $c=0$ Next note that $b = 3a$ since $b<5$ and $a>0$ the only solution is $b=3$,$a=1$ Thus in base 10 the number is $810+30=840$

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=VEbEouF2D0g&t=0s