Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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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>. | ||
− | '''Case 3:''' <math> | + | '''Case 3:''' <math>2200 \le n \le 2999</math> |
− | Let <math> | + | Let <math>2 \le k \le 9</math> be the 2nd digit of <math>n</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>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> | ||
Line 41: | Line 41: | ||
<math>(k-1)100+93 \le n-2007 \le (k-1)100+92</math>, and <math>4+k^2 \le S(n) \le 166+k^2</math> | <math>(k-1)100+93 \le n-2007 \le (k-1)100+92</math>, and <math>4+k^2 \le S(n) \le 166+k^2</math> | ||
+ | At <math>k=2</math>, <math>100(k-1)+93=193>166+k^2>170</math>. | ||
+ | At <math>k=3</math>, <math>100(k-1)+93=293>166+k^2>175</math>. | ||
+ | At <math>k=4</math>, <math>100(k-1)+93=393>166+k^2>182</math>. | ||
+ | At <math>k=5</math>, <math>100(k-1)+93=493>166+k^2>191</math>. | ||
+ | At <math>k=6</math>, <math>100(k-1)+93=593>166+k^2>202</math>. | ||
+ | At <math>k=7</math>, <math>100(k-1)+93=693>166+k^2>215</math>. | ||
+ | At <math>k=8</math>, <math>100(k-1)+93=793>166+k^2>230</math>. | ||
+ | At <math>k=9</math>, <math>100(k-1)+93=893>166+k^2>247</math>. | ||
Revision as of 15:45, 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
,
.
...ongoing writing of solution...
~Tomas Diaz. orders@tomasdiaz.com