Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
Line 97: | Line 97: | ||
<math>n-2007=10a+b-7 \le S(n)=4+a^2+b^2</math> | <math>n-2007=10a+b-7 \le S(n)=4+a^2+b^2</math> | ||
+ | |||
+ | Solving the inequality <math>10a+b-7 \le 4+a^2+b^2</math> we have: | ||
+ | |||
+ | <math>0 \le b^2-b+(a^2-10a+11)</math> | ||
+ | |||
Revision as of 16:57, 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.
NOTE... case 4 is wrong. Need to rewrite it
Case 4:
Let be the 3rd digit of
, and
, and
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
At ,
.
Since , for
, then
and there is no possible
when
when combined with the previous cases.
Case 5: Here we need to try each case from n=2008 to n=2109
Let and
be the 3rd and 4th digits of n respectively.
;
Solving the inequality we have:
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com