Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
Line 65: | Line 65: | ||
Let <math>2 \le k \le 9</math> be the 3rd digit of <math>n</math> | Let <math>2 \le k \le 9</math> be the 3rd digit of <math>n</math> | ||
− | <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>(k-1)10+ | + | <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=2</math>, <math>10(k-1)+3=193>85+k^2>89</math>. | At <math>k=2</math>, <math>10(k-1)+3=193>85+k^2>89</math>. |
Revision as of 16:02, 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 ,
.
Since , for
, then
and there is no possible
when
when combined with the previous cases.
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com