Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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NOTE... case 4 is wrong. Need to rewrite it | NOTE... case 4 is wrong. Need to rewrite it | ||
− | '''Case 4:''' <math> | + | '''Case 4:''' <math>2110 \le n \le 2199</math> |
− | Let <math> | + | Let <math>1 \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+k^2+9^2</math> | <math>2100+10k \le n \le 2109+10k</math>, and <math>2^2++1+k^2 \le S(n) \le 2^2+1+k^2+9^2</math> | ||
Line 69: | Line 69: | ||
<math>(k-1)10+93 \le n-2007 \le (k-1)10+102</math>, and <math>5+k^2 \le S(n) \le 86+k^2</math> | <math>(k-1)10+93 \le n-2007 \le (k-1)10+102</math>, and <math>5+k^2 \le S(n) \le 86+k^2</math> | ||
− | At <math>k= | + | At <math>k=1</math>, <math>10(k-1)+93=93>86+k^2>87</math>. |
− | At <math>k= | + | At <math>k=2</math>, <math>10(k-1)+93=103>86+k^2>90</math>. |
− | At <math>k= | + | At <math>k=3</math>, <math>10(k-1)+93=113>86+k^2>95</math>. |
− | At <math>k= | + | At <math>k=4</math>, <math>10(k-1)+93=123>86+k^2>102</math>. |
− | At <math>k= | + | At <math>k=5</math>, <math>10(k-1)+93=133>86+k^2>111</math>. |
− | At <math>k= | + | At <math>k=6</math>, <math>10(k-1)+93=143>86+k^2>122</math>. |
− | At <math>k= | + | At <math>k=7</math>, <math>10(k-1)+93=153>86+k^2>135</math>. |
− | At <math>k= | + | At <math>k=8</math>, <math>10(k-1)+93=163>86+k^2>150</math>. |
− | + | At <math>k=9</math>, <math>10(k-1)+93=173>86+k^2>167</math>. | |
+ | 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. | ||
+ | |||
+ | '''Case 5:''' | ||
...ongoing writing of solution... | ...ongoing writing of solution... | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com |
Revision as of 16:06, 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:
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com