Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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==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 |
Revision as of 15:07, 24 November 2023
Problem
Let be the sum of the squares of the digits of
. How many positive integers
satisfy the inequality
?
Solution
We start by rearranging the inequality the following way:
and compare the possible values for the left hand side and the right hand side of this innequality.
Case 1: has 5 digits or more.
Let = number of digits of n.
Then as a function of d,
, and
, and
when ,
Since for
, then
and there is no
possible when
has 5 or more digits.
~Tomas Diaz. orders@tomasdiaz.com