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Difference between revisions of "2007 iTest Problems/Problem 35"

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==Solution==
 
==Solution==
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We seek to maximize the number of digits in this number (call it <math>N</math>), which requires maximizing the arithmetic means to maximize each digit. Clearly, the first and last digits should be 9. Clearly digits cannot repeat consecutively unless they are the middle digits, and digits are descending symmetric to the middle digits (like 9862689 or 987789). We will build <math>N</math> by finding its descending sequence and then repeating it ascending to complete the number. Consider if the number has two consecutive digits in its descending sequence, like 98 or 9754. Let then be <math>n</math> and <math>n-1</math>. Then the next number <math>x</math> must satisfy <math>n+x>2(n-1)</math> or <math>x>n-2</math>. Thus the descending sequence must end, so in order to continue it we must have digits that differ by at least 2. The number with the most amount of digits satisfying this property is 97531013579, and we verify that the property in the problem is satisfied. Therefore <math>N=\boxed{97531013579}</math>.
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We seek to maximize the number of digits in the number (call it <math>N</math>) first, so the first and last digits should be 9. Next recognize that <math>N</math> should be symmetric (like 5445 or 93239), and that if there are ever consecutive digits the number cannot contain any digits less than the lower consecutive digit. If <math>N</math> begins with 98, then it must be 9889. If <math>N</math> begins with 97, it can either be 9779, 97679, or 976679. If <math>N</math> begins with 96, it can either be 9669, 96569, 965569, 96469, 964469, 9643469, or 96433469. If <math>N</math> begins with 95, it has the maximum value 95322359, which is less than 96433469. Clearly if <math>N</math> starts with a lower second digit it is less than this number, so <math>N=\boxed{96433469}</math>. Note the differences between consecutive digits in this number.
  
 
==See Also==
 
==See Also==

Latest revision as of 22:09, 7 April 2024

Problem 35

Find the greatest natural number possessing the property that each of its digits except the first and last one is less than the arithmetic mean of the two neighboring digits.

Solution

We seek to maximize the number of digits in the number (call it $N$) first, so the first and last digits should be 9. Next recognize that $N$ should be symmetric (like 5445 or 93239), and that if there are ever consecutive digits the number cannot contain any digits less than the lower consecutive digit. If $N$ begins with 98, then it must be 9889. If $N$ begins with 97, it can either be 9779, 97679, or 976679. If $N$ begins with 96, it can either be 9669, 96569, 965569, 96469, 964469, 9643469, or 96433469. If $N$ begins with 95, it has the maximum value 95322359, which is less than 96433469. Clearly if $N$ starts with a lower second digit it is less than this number, so $N=\boxed{96433469}$. Note the differences between consecutive digits in this number.

See Also

2007 iTest (Problems, Answer Key)
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Problem 34 		Followed by:
Problem 36
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