Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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'''Case 5:''' <math>2008 \le n \le 2099</math> | '''Case 5:''' <math>2008 \le n \le 2099</math> | ||
− | Let <math>a</math> and <math>b</math> be the 3rd and 4th digits of n respectively | + | Let <math>a</math> and <math>b</math> be the 3rd and 4th digits of n respectively with <math>0 \le a \le 9</math> and <math>0 \le b \le 9</math> |
<math>n=2000+10a+b</math>; <math>S(n)=4+a^2+b^2</math> | <math>n=2000+10a+b</math>; <math>S(n)=4+a^2+b^2</math> | ||
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<math>0 \le b^2-b+(a^2-10a+11)</math> | <math>0 \le b^2-b+(a^2-10a+11)</math> | ||
− | When <math>a=0\;</math>, <math>0 \le b^2-b+11</math>, which gives: <math>b \ge 0</math>. Which is <math>n=2008</math> and <math>n=2009</math> Total possible <math>n</math>'s: '''2''' | + | When <math>a=0\;</math>, <math>0 \le b^2-b+11</math>, which gives: <math>9 \ge b \ge 0</math>. Which is <math>n=2008</math> and <math>n=2009</math> Total possible <math>n</math>'s: '''2''' |
− | When <math>a=1</math>, <math>0 \le b^2-b+2</math>, which gives: <math>b \ge 0</math>. Total possible <math>n</math>'s: '''10''' | + | When <math>a=1</math>, <math>0 \le b^2-b+2</math>, which gives: <math>9 \ge b \ge 0</math>. Total possible <math>n</math>'s: '''10''' |
− | When <math>a=2</math>, <math>0 \le b^2-b-5</math>, which gives: <math>b \ge 3</math>. Total possible <math>n</math>'s: '''7''' | + | When <math>a=2</math>, <math>0 \le b^2-b-5</math>, which gives: <math>9 \ge b \ge 3</math>. Total possible <math>n</math>'s: '''7''' |
− | When <math>a=3</math>, <math>0 \le b^2-b-10</math>, which gives: <math>b \ge 4</math>. Total possible <math>n</math>'s: '''6''' | + | When <math>a=3</math>, <math>0 \le b^2-b-10</math>, which gives: <math>9 \ge b \ge 4</math>. Total possible <math>n</math>'s: '''6''' |
− | When <math>a=4</math>, <math>0 \le b^2-b-13</math>, which gives: <math>b \ge 5</math>. Total possible <math>n</math>'s: '''5''' | + | When <math>a=4</math>, <math>0 \le b^2-b-13</math>, which gives: <math>9 \ge b \ge 5</math>. Total possible <math>n</math>'s: '''5''' |
− | When <math>a=5</math>, <math>0 \le b^2-b-14</math>, which gives: <math>b \ge 5</math>. Total possible <math>n</math>'s: '''5''' | + | When <math>a=5</math>, <math>0 \le b^2-b-14</math>, which gives: <math>9 \ge b \ge 5</math>. Total possible <math>n</math>'s: '''5''' |
− | When <math>a=6</math>, <math>0 \le b^2-b-13</math>, which gives: <math>b \ge 5</math>. Total possible <math>n</math>'s: '''5''' | + | When <math>a=6</math>, <math>0 \le b^2-b-13</math>, which gives: <math>9 \ge b \ge 5</math>. Total possible <math>n</math>'s: '''5''' |
− | When <math>a=7</math>, <math>0 \le b^2-b-10</math>, which gives: <math>b \ge 4</math>. Total possible <math>n</math>'s: '''6''' | + | When <math>a=7</math>, <math>0 \le b^2-b-10</math>, which gives: <math>9 \ge b \ge 4</math>. Total possible <math>n</math>'s: '''6''' |
− | When <math>a=8</math>, <math>0 \le b^2-b-5</math>, which gives: <math>b \ge 3</math>. Total possible <math>n</math>'s: '''7''' | + | When <math>a=8</math>, <math>0 \le b^2-b-5</math>, which gives: <math>9 \ge b \ge 3</math>. Total possible <math>n</math>'s: '''7''' |
− | When <math>a=9</math>, <math>0 \le b^2-b+2</math>, which gives: <math>b \ge 0</math>. Total possible <math>n</math>'s: '''10''' | + | When <math>a=9</math>, <math>0 \le b^2-b+2</math>, which gives: <math>9 \ge b \ge 0</math>. Total possible <math>n</math>'s: '''10''' |
− | |||
− | |||
Therefore, the total number of possible <math>n</math>'s is: <math>2+10+7+6+5+5+5+6+7+10=\boxed{63}</math> | Therefore, the total number of possible <math>n</math>'s is: <math>2+10+7+6+5+5+5+6+7+10=\boxed{63}</math> |
Revision as of 22:49, 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 ,
,
.
At ,
,
.
At ,
,
.
Since , for
, then
and there is no possible
when
when combined with the previous cases.
From cases 1 through 4 we now know that
Case 5:
Let and
be the 3rd and 4th digits of n respectively with
and
;
Solving the inequality we have:
When ,
, which gives:
. Which is
and
Total possible
's: 2
When ,
, which gives:
. Total possible
's: 10
When ,
, which gives:
. Total possible
's: 7
When ,
, which gives:
. Total possible
's: 6
When ,
, which gives:
. Total possible
's: 5
When ,
, which gives:
. Total possible
's: 5
When ,
, which gives:
. Total possible
's: 5
When ,
, which gives:
. Total possible
's: 6
When ,
, which gives:
. Total possible
's: 7
When ,
, which gives:
. Total possible
's: 10
Therefore, the total number of possible 's is:
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.