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

(Created page with "==Problem 2== 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}\u...")
 
 
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Next note that <math>b = 3a</math> since <math>b<5</math> and <math>a>0</math> the only solution is <math>b=3</math>,<math>a=1</math>
 
Next note that <math>b = 3a</math> since <math>b<5</math> and <math>a>0</math> the only solution is <math>b=3</math>,<math>a=1</math>
 
Thus in base 10 the number is <math>810+30=840</math>
 
Thus in base 10 the number is <math>810+30=840</math>
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==Video Solution by Punxsutawney Phil==
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https://www.youtube.com/watch?v=VEbEouF2D0g&t=0s

Latest revision as of 15:06, 19 January 2021

Problem 2

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