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...") |
Rockmanex3 (talk | contribs) (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>. | ||
− | == Solution == | + | ==Solution== |
+ | |||
+ | Using the definition of [[base numbers]], the equation can be rewritten as | ||
+ | <cmath>B^2 + B + 1 = 9072a + 252b + 7c</cmath> | ||
+ | <cmath>\frac{B^2 + B + 1}{7} = 1296a + 36b + c</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | * 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>. | ||
+ | |||
+ | * 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>. | ||
+ | |||
+ | * 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>. | ||
+ | |||
+ | * 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>. | ||
+ | |||
+ | * 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>. | ||
+ | |||
+ | * 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{iTest box|year=2007|num-b=44|num-a=46}} | ||
+ | |||
+ | [[Category:Olympiad Number Theory Problems]] |
Revision as of 05:05, 16 June 2018
Problem
Find the sum of all positive integers such that , where represent distinct base digits, .
Solution
Using the definition of base numbers, the equation can be rewritten as
To find the values of , use casework for values of since has the most influence on the value of . Casework will be heavy, but a few tips can lighten the load. First, since is one less than a multiple of , is congruent to or modulo . Second, once , can not be higher for a given . Third, use estimation to approximate the lower bound for a given .
- If , then is just more than . The first few values of that work are , , , and . Testing each case, when and .
- If , then is just more than . The first few values of that work are , , , and . Testing each case, when and .
- If , then is just more than . The first few values of that work are , , . Testing each case, when and .
- If , then is just more than . The first few values of that work are and . After testing each case, no values of work when .
- If , then is just more than . The first few values of that work are and . After testing each case, once again, no values of work when .
- If , then is just more than . The first few values of that work are and . After testing each case, yet again, no values of work when .
In summary, the only possible values of are and , and the sum of the values equals .
See Also
2007 iTest (Problems) | ||
Preceded by: Problem 44 |
Followed by: Problem 46 | |
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