Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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<math>2000+100k \le n \le 2099+100k</math>, and <math>2^2+k^2 \le S(n) \le 2^2+k^2+2 \times 9^2</math> | <math>2000+100k \le n \le 2099+100k</math>, and <math>2^2+k^2 \le S(n) \le 2^2+k^2+2 \times 9^2</math> | ||
− | <math> | + | <math>100k-7 \le n-2007 \le 100k+92</math>, and <math>4+k^2 \le S(n) \le 166+k^2</math> |
− | At <math>k=2</math>, <math> | + | At <math>k=2</math>, <math>100k-7=193\;\;86+k^2=170</math>, <math>193>170</math>. |
− | At <math>k=3</math>, <math> | + | At <math>k=3</math>, <math>100k-7=293\;\;86+k^2=175</math>, <math>293>175</math>. |
− | At <math>k=4</math>, <math> | + | At <math>k=4</math>, <math>100k-7=393\;\;86+k^2=182</math>, <math>393>182</math>. |
− | At <math>k=5</math>, <math> | + | At <math>k=5</math>, <math>100k-7=493\;\;86+k^2=191</math>, <math>493>191</math>. |
− | At <math>k=6</math>, <math> | + | At <math>k=6</math>, <math>100k-7=593\;\;86+k^2=202</math>, <math>593>202</math>. |
− | At <math>k=7</math>, <math> | + | At <math>k=7</math>, <math>100k-7=693\;\;86+k^2=215</math>, <math>693>215</math>. |
− | At <math>k=8</math>, <math> | + | At <math>k=8</math>, <math>100k-7=793\;\;86+k^2=230</math>, <math>793>230</math>. |
− | At <math>k=9</math>, <math> | + | At <math>k=9</math>, <math>100k-7=893\;\;86+k^2=247</math>, <math>893>247</math>. |
− | Since <math> | + | Since <math>100k-7 > 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. |
Revision as of 22:30, 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.