Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"

Revision as of 15:07, 24 November 2023

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

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 innequality.

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 $1 \le S(n) \le 9^2d$

$10^d - 2007 \le n < 10^{d+1}-2008$ , 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 $S(n)$ possible when $n$ has 5 or more digits.

~Tomas Diaz. orders@tomasdiaz.com

@@ Line 3: / Line 3: @@
 ==Solution==
-{{Solution}}
+We start by rearranging the inequality the following way:
+<math>n-2007\le S(n)</math> and compare the possible values for the left hand side and the right hand side of this innequality.
+'''Case 1:''' <math>n</math> has 5 digits or more.
+Let <math>d</math> = number of digits of n.
+Then as a function of d,
+<math>10^d \le n < 10^{d+1}-1</math>, and <math>1 \le S(n) \le 9^2d</math>
+<math>10^d - 2007 \le n < 10^{d+1}-2008</math>, and <math>1 \le S(n) \le 81d</math>
+when <math>d \ge 5</math>,
+<math>10^d - 2007 \ge 10^5 -2007</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 <math>S(n)</math> possible when <math>n</math> has 5 or more digits.
+~Tomas Diaz. orders@tomasdiaz.com