document.write( "Question 1204005: A Spooky number is a number which is 13 times the sum of its digits.
\n" ); document.write( "Find all Spooky numbers (and prove that there aren’t any more).
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Algebra.Com's Answer #839982 by math_tutor2020(3835)\"\" \"About 
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\n" ); document.write( "One digit numbers:\r
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\n" ); document.write( "\n" ); document.write( "x = some positive integer in the set {0,1,2,3,...,8,9}
\n" ); document.write( "If x was \"spooky\", then x = 13x which solves to x = 0
\n" ); document.write( "Therefore, 0 is the only spooky 1 digit number. \r
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\n" ); document.write( "\n" ); document.write( "Two digit numbers:\r
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\n" ); document.write( "\n" ); document.write( "Consider the 2 digit number 10a+b, where a,b are positive integers.
\n" ); document.write( "Eg: a = 2 and b = 3 gives 10a+b = 23.\r
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\n" ); document.write( "\n" ); document.write( "If a spooky number could be 2 digits, then this equation would be true
\n" ); document.write( "10a+b = 13*(a+b)\r
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\n" ); document.write( "\n" ); document.write( "Solve for 'a' to get a = -4b. \r
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\n" ); document.write( "\n" ); document.write( "If b > 0, then a < 0.
\n" ); document.write( "This lets us conclude that there aren't any 2 digit spooky numbers that are positive.\r
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\n" ); document.write( "\n" ); document.write( "Furthermore, a+b > 0 so 13*(a+b) > 0. This means spooky numbers cannot be negative.\r
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\n" ); document.write( "\n" ); document.write( "Conclusion for this section: There aren't any spooky 2 digit numbers.\r
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\n" ); document.write( "\n" ); document.write( "Three digit numbers:\r
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\n" ); document.write( "\n" ); document.write( "a,b,c are positive integers
\n" ); document.write( "100a+10b+c is some 3 digit number
\n" ); document.write( "100a+10b+c = 13*(a+b+c) must be the case if this number is considered \"spooky\".\r
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\n" ); document.write( "\n" ); document.write( "Solve for 'b' to get b = 29a-4c
\n" ); document.write( "Plug that into 100a+10b+c to get...
\n" ); document.write( "100a+10b+c
\n" ); document.write( "100a+10(29a-4c)+c
\n" ); document.write( "100a+290a-40c+c
\n" ); document.write( "390a-39c
\n" ); document.write( "39*(10a-c)\r
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\n" ); document.write( "\n" ); document.write( "This shows 100a+10b+c is a multiple of 39 if 100a+10b+c is spooky.\r
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\n" ); document.write( "\n" ); document.write( "Let's look at 13*(a+b+c) more closely.
\n" ); document.write( "Specifically a+b+c\r
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\n" ); document.write( "\n" ); document.write( "The smallest that sum can be is 1 because of the number 100
\n" ); document.write( "The largest is 9+9+9 = 27\r
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\n" ); document.write( "\n" ); document.write( "1 <= a+b+c <= 27
\n" ); document.write( "or
\n" ); document.write( "1 <= a+b+c <= 9*3\r
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\n" ); document.write( "\n" ); document.write( "Multiply both sides by 13 to get
\n" ); document.write( "1 <= a+b+c <= 27
\n" ); document.write( "13*1 <= 13*(a+b+c) <= 13*27
\n" ); document.write( "13 <= 13*(a+b+c) <= 351\r
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\n" ); document.write( "\n" ); document.write( "It appears that 3 digit spooky numbers could be possible.
\n" ); document.write( "I'll let the student search for them. \r
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\n" ); document.write( "\n" ); document.write( "Four digit numbers:\r
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\n" ); document.write( "\n" ); document.write( "We'll do a similar trick as the previous section.
\n" ); document.write( "1 <= a+b+c+d <= 9+9+9+9
\n" ); document.write( "1 <= a+b+c+d <= 9*4
\n" ); document.write( "1 <= a+b+c+d <= 36
\n" ); document.write( "13*1 <= 13*(a+b+c+d) <= 13*36
\n" ); document.write( "13 <= 13*(a+b+c+d) <= 468\r
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\n" ); document.write( "\n" ); document.write( "We run into a problem. The largest 13*(a+b+c+d) can be is 468 which is clearly NOT a 4 digit number.
\n" ); document.write( "There's no way to have 13*(a+b+c+d) as a 4 digit number.\r
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\n" ); document.write( "\n" ); document.write( "Therefore, we conclude that 4 digit spooky numbers aren't possible.\r
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\n" ); document.write( "\n" ); document.write( "The logic for 5 digit numbers, 6 digits, and so on will follow the same idea. I'll let the student finish the proof.
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