Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
Line 63: | Line 63: | ||
− | '''Case 4:''' <math> | + | '''Case 4:''' <math>2100 \le n \le 2199</math> |
− | Let <math> | + | Let <math>0 \le k \le 9</math> be the 3rd digit of <math>n</math> |
− | <math>2100+10k \le n \le 2109+10k</math>, and <math>2^2+1+k^2 \le S(n) \le 2^2+1^2+k^2+9^2</math> | + | <math>2100+10k \le n \le 2109+10k</math>, and <math>2^2+1^2+k^2 \le S(n) \le 2^2+1^2+k^2+9^2</math> |
− | <math>10k | + | <math>10k+93 \le n-2007 \le 10k+102</math>, and <math>5+k^2 \le S(n) \le 86+k^2</math> |
− | At <math>k= | + | At <math>k=0</math>, <math>10k+93=93\;and\;86+k^2=86</math>, <math>93>86</math>. |
− | At <math>k= | + | At <math>k=1</math>, <math>10k+93=103\;and\;86+k^2=87</math>, <math>103>87</math>. |
− | At <math>k= | + | At <math>k=2</math>, <math>10k+93=113\;and\;86+k^2=90</math>, <math>113>90</math>. |
− | At <math>k= | + | At <math>k=3</math>, <math>10k+93=123\;and\;86+k^2=95</math>, <math>123>95</math>. |
− | At <math>k= | + | At <math>k=4</math>, <math>10k+93=133\;and\;86+k^2=102</math>, <math>133>102</math>. |
− | At <math>k= | + | At <math>k=5</math>, <math>10k+93=143\;and\;86+k^2=111</math>, <math>143>111</math>. |
− | At <math>k= | + | At <math>k=6</math>, <math>10k+93=153\;and\;86+k^2=122</math>, <math>153>122</math>. |
− | At <math>k= | + | At <math>k=7</math>, <math>10k+93=163\;and\;86+k^2=135</math>, <math>163>135</math>. |
− | At <math>k= | + | At <math>k=8</math>, <math>10k+93=173\;and\;86+k^2=150</math>, <math>173>150</math>. |
− | + | At <math>k=9</math>, <math>10k+93=183\;and\;86+k^2=167</math>, <math>183>167</math>. | |
+ | Since <math>10k+93 > 85+k^2</math>, for <math>0 \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 2100</math> when combined with the previous cases. | ||
− | '''Case 5:''' | + | From cases 1 through 4 we now know that <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. |
Revision as of 22:47, 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.
;
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.