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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
Line 23: Line 23:
 
<math>10^d - 2007 \ge 10^5 -2007 > 81d</math>
 
<math>10^d - 2007 \ge 10^5 -2007 > 81d</math>
  
−
Since <math>10^d - 2007 > 81d</math> for <math>d \ge 5</math>, then <math>n-2007\not\le S(n)</math> and there is no possible <math>n</math> when <math>n</math> has 5 or more digits.
+
Since <math>10^d - 2007 > 81d</math> for <math>d \ge 5</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 5 or more digits.
  
 
'''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math>
 
'''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math>
Line 31: Line 31:
 
<math>993 \le n-2007 \le 7992</math>, and <math>9 \le S(n) \le 252</math>
 
<math>993 \le n-2007 \le 7992</math>, and <math>9 \le S(n) \le 252</math>
  
−
Since <math>993 > 252</math>, then <math>n-2007\not\le S(n)</math> and there is no possible <math>n</math> when <math>n</math> has 4 digits and <math>n \ge 3000</math>.
+
Since <math>993 > 252</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 4 digits and <math>n \ge 3000</math>.
  
  

Revision as of 14:20, 24 November 2023

Problem

Let $S(n)$ be the sum of the squares of the digits of $n$. How many positive integers $n>2007$ satisfy the inequality $n-S(n)\le 2007$?

Solution

We start by rearranging the inequality the following way:

$n-2007\le S(n)$ and compare the possible values for the left hand side and the right hand side of this inequality.

Case 1: $n$ has 5 digits or more.

Let $d$ = number of digits of n.

Then as a function of d,

$10^d \le n < 10^{d+1}-1$, and $1 \le S(n) \le 9^2d$

$10^d - 2007 \le n-2007 < 10^{d+1}-2008$, and $1 \le S(n) \le 81d$

when $d \ge 5$,

$10^d - 2007 \ge 10^5 -2007$

$10^d - 2007 \ge 10^5 -2007 > 81d$

Since $10^d - 2007 > 81d$ for $d \ge 5$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n$ has 5 or more digits.

Case 2: $n$ has 4 digits and $n \ge 3000$

$3000 \le n \le 9999$, and $3^2 \le S(n) \le 3^2+3 \times 9^2$

$993 \le n-2007 \le 7992$, and $9 \le S(n) \le 252$

Since $993 > 252$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n$ has 4 digits and $n \ge 3000$.


...ongoing writing of solution...

~Tomas Diaz. orders@tomasdiaz.com

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