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Difference between revisions of "2007 iTest Problems/Problem 45"

(Created page with "== Problem == Find the sum of all positive integers <math>B</math> such that <math>(111)_B=(aabbcc)_6</math>, where <math>a,b,c</math> represent distinct base <math>6</math> dig...")
 
(Solution to Problem 45 — grinding the base numbers)
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Find the sum of all positive integers <math>B</math> such that <math>(111)_B=(aabbcc)_6</math>, where <math>a,b,c</math> represent distinct base <math>6</math> digits, <math>a\neq 0</math>.  
 
Find the sum of all positive integers <math>B</math> such that <math>(111)_B=(aabbcc)_6</math>, where <math>a,b,c</math> represent distinct base <math>6</math> digits, <math>a\neq 0</math>.  
  
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== Solution ==
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==Solution==
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Using the definition of [[base numbers]], the equation can be rewritten as
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<cmath>B^2 + B + 1 = 9072a + 252b + 7c</cmath>
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<cmath>\frac{B^2 + B + 1}{7} = 1296a + 36b + c</cmath>
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To find the values of <math>B</math>, use [[casework]] for values of <math>a</math> since <math>a</math> has the most influence on the value of <math>7(1296a + 36b + c) - 1</math>.  Casework will be heavy, but a few tips can lighten the load.  First, since <math>B^2 + B</math> is one less than a multiple of <math>7</math>, <math>B</math> is congruent to <math>2</math> or <math>4</math> [[modulo]] <math>7</math>.  Second, once <math>\frac{B^2 + B + 1}{7} > 1296a + 185</math>, <math>B</math> can not be higher for a given <math>a</math>.  Third, use [[estimation]] to approximate the lower bound for a given <math>a</math>.
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* If <math>a = 1</math>, then <math>B^2 + B</math> is just more than <math>9072</math>.  The first few values of <math>B</math> that work are <math>95</math>, <math>100</math>, <math>102</math>, and <math>107</math>.  Testing each case, <math>B = 100</math> when <math>b = 4</math> and <math>c = 3</math>.
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* If <math>a = 1</math>, then <math>B^2 + B</math> is just more than <math>9072</math>.  The first few values of <math>B</math> that work are <math>95</math>, <math>100</math>, <math>102</math>, and <math>107</math>.  Testing each case, <math>B = 100</math> when <math>b = 4</math> and <math>c = 3</math>.
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* If <math>a = 2</math>, then <math>B^2 + B</math> is just more than <math>18144</math>.  The first few values of <math>B</math> that work are <math>137</math>, <math>142</math>, <math>144</math>.  Testing each case, <math>B = 137</math> when <math>b = 3</math> and <math>c = 1</math>.
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* If <math>a = 3</math>, then <math>B^2 + B</math> is just more than <math>27216</math>.  The first few values of <math>B</math> that work are <math>165</math> and <math>170</math>.  After testing each case, no values of <math>B</math> work when <math>a = 3</math>.
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* If <math>a = 4</math>, then <math>B^2 + B</math> is just more than <math>36288</math>.  The first few values of <math>B</math> that work are <math>191</math> and <math>193</math>.  After testing each case, once again, no values of <math>B</math> work when <math>a = 4</math>.
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* If <math>a = 5</math>, then <math>B^2 + B</math> is just more than <math>45360</math>.  The first few values of <math>B</math> that work are <math>214</math> and <math>219</math>.  After testing each case, yet again, no values of <math>B</math> work when <math>a = 5</math>.
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In summary, the only possible values of <math>B</math> are <math>100</math> and <math>137</math>, and the sum of the values equals <math>\boxed{237}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=44|num-a=46}}
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[[Category:Olympiad Number Theory Problems]]

Revision as of 05:05, 16 June 2018

Problem

Find the sum of all positive integers $B$ such that $(111)_B=(aabbcc)_6$, where $a,b,c$ represent distinct base $6$ digits, $a\neq 0$.

Solution

Using the definition of base numbers, the equation can be rewritten as \[B^2 + B + 1 = 9072a + 252b + 7c\] \[\frac{B^2 + B + 1}{7} = 1296a + 36b + c\]

To find the values of $B$, use casework for values of $a$ since $a$ has the most influence on the value of $7(1296a + 36b + c) - 1$. Casework will be heavy, but a few tips can lighten the load. First, since $B^2 + B$ is one less than a multiple of $7$, $B$ is congruent to $2$ or $4$ modulo $7$. Second, once $\frac{B^2 + B + 1}{7} > 1296a + 185$, $B$ can not be higher for a given $a$. Third, use estimation to approximate the lower bound for a given $a$.

  • If $a = 1$, then $B^2 + B$ is just more than $9072$. The first few values of $B$ that work are $95$, $100$, $102$, and $107$. Testing each case, $B = 100$ when $b = 4$ and $c = 3$.
  • If $a = 1$, then $B^2 + B$ is just more than $9072$. The first few values of $B$ that work are $95$, $100$, $102$, and $107$. Testing each case, $B = 100$ when $b = 4$ and $c = 3$.
  • If $a = 2$, then $B^2 + B$ is just more than $18144$. The first few values of $B$ that work are $137$, $142$, $144$. Testing each case, $B = 137$ when $b = 3$ and $c = 1$.
  • If $a = 3$, then $B^2 + B$ is just more than $27216$. The first few values of $B$ that work are $165$ and $170$. After testing each case, no values of $B$ work when $a = 3$.
  • If $a = 4$, then $B^2 + B$ is just more than $36288$. The first few values of $B$ that work are $191$ and $193$. After testing each case, once again, no values of $B$ work when $a = 4$.
  • If $a = 5$, then $B^2 + B$ is just more than $45360$. The first few values of $B$ that work are $214$ and $219$. After testing each case, yet again, no values of $B$ work when $a = 5$.

In summary, the only possible values of $B$ are $100$ and $137$, and the sum of the values equals $\boxed{237}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 44 		Followed by:
Problem 46
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