Difference between revisions of "2021 CIME I Problems/Problem 2"
Dark adonis (talk | contribs) (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> | ||
| + | |||
| + | ==Video Solution by Punxsutawney Phil== | ||
| + | https://www.youtube.com/watch?v=VEbEouF2D0g&t=0s | ||
Revision as of 14:06, 19 January 2021
Problem 2
For digits 
with 
the positive integer 
can be written as 
in base 
and 
in base 
. Find the base-
representation of .
Solution
Consider the different representations of the number and equate them:
![\[(9^3 + 9^2) a +(9+1)b = (5^4+5^3) a + (5^2+5)b+c\]](http://latex.artofproblemsolving.com/6/f/3/6f32db2720a04625c6676044f21293d23315f6c7.png)
![\[(810)a+10b = (750)a+30b+c\]](http://latex.artofproblemsolving.com/2/f/6/2f6bce0a74e628963a0c5111e8b10e489a100d14.png)
![\[60 a - 20b-c=0\]](http://latex.artofproblemsolving.com/d/9/8/d98f66f16c46b3e51c652be3e281a3fdb9ef7777.png)
Note that c can't contribute since it is less than 5 so 
Next note that 
since 
and 
the only solution is 
,
Thus in base 10 the number is 
