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

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==Solution==
 
==Solution==
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 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 them 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>.
  
 
==See Also==
 
==See Also==

Revision as of 00:16, 6 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 this number (call it $N$), 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 $N$ 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 them be $n$ and $n-1$. Then the next number $x$ must satisfy $n+x>2(n-1)$ or $x>n-2$. 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 $N=\boxed{97531013579}$.

See Also

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