Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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<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> | ||
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<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 15:20, 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 inequality.
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.
Case 2: has 4 digits and
, and
, and
Since , then
and there is no possible
when
has 4 digits and
.
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com