Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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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 2008 and 2009 Total | + | 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=1</math>, <math>0 \le b^2-b+2</math>, which gives: <math>b \ge 0</math>. Total | + | 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=2</math>, <math>0 \le b^2-b-5</math>, which gives: <math>b \ge 3</math>. Total | + | 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=3</math>, <math>0 \le b^2-b-10</math>, which gives: <math>b \ge 4</math>. Total | + | 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=4</math>, <math>0 \le b^2-b-13</math>, which gives: <math>b \ge 5</math>. Total | + | 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=5</math>, <math>0 \le b^2-b-14</math>, which gives: <math>b \ge 5</math>. Total | + | 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=6</math>, <math>0 \le b^2-b-13</math>, which gives: <math>b \ge 5</math>. Total | + | 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=7</math>, <math>0 \le b^2-b-10</math>, which gives: <math>b \ge 4</math>. Total | + | 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=8</math>, <math>0 \le b^2-b-5</math>, which gives: <math>b \ge 3</math>. Total | + | 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=9</math>, <math>0 \le b^2-b+2</math>, which gives: <math>b \ge 0</math>. Total | + | 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 |
No valid <math>n</math> for <math>2100 le n le 2109</math> | No valid <math>n</math> for <math>2100 le n le 2109</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> | |
...ongoing writing of solution... | ...ongoing writing of solution... | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com |
Revision as of 17:02, 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:
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
No valid for
Therefore, the total number of possible 's is:
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com