Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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<math>n-2007\le S(n)</math> and compare the possible values for the left hand side and the right hand side of this inequality. | <math>n-2007\le S(n)</math> and compare the possible values for the left hand side and the right hand side of this inequality. | ||
+ | |||
'''Case 1:''' <math>n</math> has 5 digits or more. | '''Case 1:''' <math>n</math> has 5 digits or more. | ||
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Since <math>10^d - 2007 > 81d</math> for <math>d \ge 5</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 5 or more digits. | Since <math>10^d - 2007 > 81d</math> for <math>d \ge 5</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 5 or more digits. | ||
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'''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math> | '''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math> | ||
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Since <math>993 > 252</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 4 digits and <math>n \ge 3000</math>. | Since <math>993 > 252</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 4 digits and <math>n \ge 3000</math>. | ||
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'''Case 3:''' <math>2200 \le n \le 2999</math> | '''Case 3:''' <math>2200 \le n \le 2999</math> | ||
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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. | ||
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'''Case 4:''' <math>2110 \le n \le 2199</math> | '''Case 4:''' <math>2110 \le n \le 2199</math> | ||
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Since <math>10(k-1)+93 > 85+k^2</math>, for <math>1 \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 2110</math> when combined with the previous cases. | Since <math>10(k-1)+93 > 85+k^2</math>, for <math>1 \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 2110</math> when combined with the previous cases. | ||
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'''Case 5:''' | '''Case 5:''' |
Revision as of 22:13, 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 ,
.
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:
~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.