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

Revision as of 15:16, 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 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 $1 \le S(n) \le 9^2d$

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

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

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

...ongoing writing of solution...

~Tomas Diaz. orders@tomasdiaz.com

@@ Line 26: / Line 26: @@
 '''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math>
+<math>3000 \le n < 9999</math>, and <math>3^2 \le S(n) \le 3^2+3 \times 9^2</math>
 ...ongoing writing of solution...
 ~Tomas Diaz. orders@tomasdiaz.com