Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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Since <math>100(k-1)+93 > 166+k^2</math>, for <math>2 \le k \le 9</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n \ge 2200</math> when combined with the previous cases. | Since <math>100(k-1)+93 > 166+k^2</math>, for <math>2 \le k \le 9</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n \ge 2200</math> when combined with the previous cases. | ||
− | '''Case 4:''' <math> | + | '''Case 4:''' <math>2220 \le n \le 2299</math> |
Let <math>2 \le k \le 9</math> be the 3rd digit of <math>n</math> | Let <math>2 \le k \le 9</math> be the 3rd digit of <math>n</math> |
Revision as of 15:56, 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
.
Case 3:
Let be the 2nd digit of
, and
, and
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
Since , for
, then
and there is no possible
when
when combined with the previous cases.
Case 4:
Let be the 3rd digit of
, and
, and
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
Since , for
, then
and there is no possible
when
when combined with the previous cases.
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com