Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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Substituting for all values of a in the above inequality we get: | Substituting for all values of a in the above inequality we get: | ||
− | When <math>a=0,\; | + | When <math>a=0,\;\;0 \le b^2-b+11</math>, which gives: <math>0 \le b \le 9</math>. But <math>n>2007</math>, So, <math>n=2008</math> and <math>n=2009</math>. Total possible <math>n</math>'s: '''2''' |
− | When <math>a=1,\; | + | When <math>a=1,\;\;0 \le b^2-b+2</math>, which gives: <math>0 \le b \le 9</math>. Total possible <math>n</math>'s: '''10''' |
− | When <math>a=2,\; | + | When <math>a=2,\;\;0 \le b^2-b-5</math>, which gives: <math>3 \le b \le 9</math>. Total possible <math>n</math>'s: '''7''' |
− | When <math>a=3,\; | + | When <math>a=3,\;\;0 \le b^2-b-10</math>, which gives: <math>4 \le b \le 9</math>. Total possible <math>n</math>'s: '''6''' |
− | When <math>a=4,\; | + | When <math>a=4,\;\;0 \le b^2-b-13</math>, which gives: <math>5 \le b \le 9</math>. Total possible <math>n</math>'s: '''5''' |
− | When <math>a=5,\; | + | When <math>a=5,\;\;0 \le b^2-b-14</math>, which gives: <math>5 \le b \le 9</math>. Total possible <math>n</math>'s: '''5''' |
− | When <math>a=6,\; | + | When <math>a=6,\;\;0 \le b^2-b-13</math>, which gives: <math>5 \le b \le 9</math>. Total possible <math>n</math>'s: '''5''' |
− | When <math>a=7,\; | + | When <math>a=7,\;\;0 \le b^2-b-10</math>, which gives: <math>4 \le b \le 9</math>. Total possible <math>n</math>'s: '''6''' |
− | When <math>a=8,\; | + | When <math>a=8,\;\;0 \le b^2-b-5</math>, which gives: <math>3 \le b \le 9</math>. Total possible <math>n</math>'s: '''7''' |
− | When <math>a=9,\; | + | When <math>a=9,\;\;0 \le b^2-b+2</math>, which gives: <math>0 \le b \le 9</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> |
Latest revision as of 23:02, 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.
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:
Substituting for all values of a in the above inequality we get:
When , which gives:
. But
, So,
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.