Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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<math>n-2007\le S(n)</math> and compare the possible values for the left hand side and the right hand side of this inequality. | <math>n-2007\le S(n)</math> and compare the possible values for the left hand side and the right hand side of this inequality. | ||
+ | |||
'''Case 1:''' <math>n</math> has 5 digits or more. | '''Case 1:''' <math>n</math> has 5 digits or more. | ||
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Since <math>10^d - 2007 > 81d</math> for <math>d \ge 5</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 5 or more digits. | Since <math>10^d - 2007 > 81d</math> for <math>d \ge 5</math>, then <math>n-2007\not\le S(n)</math> and there is '''no possible <math>n</math>''' when <math>n</math> has 5 or more digits. | ||
+ | |||
'''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math> | '''Case 2:''' <math>n</math> has 4 digits and <math>n \ge 3000</math> | ||
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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>2200 \le n \le 2999</math> | '''Case 3:''' <math>2200 \le n \le 2999</math> | ||
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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>100k-7=193\;\;86+k^2=170</math>, <math>193>170</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>100k-7=393\;\;86+k^2=182</math>, <math>393>182</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>100k-7=593\;\;86+k^2=202</math>, <math>593>202</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>100k-7=793\;\;86+k^2=230</math>, <math>793>230</math>. | ||
− | At <math>k= | + | At <math>k=9</math>, <math>100k-7=893\;\;86+k^2=247</math>, <math>893>247</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. | ||
− | ... | + | |
+ | '''Case 4:''' <math>2100 \le n \le 2199</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+1^2+k^2 \le S(n) \le 2^2+1^2+k^2+9^2</math> | ||
+ | |||
+ | <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=0</math>, <math>10k+93=93\;and\;86+k^2=86</math>, <math>93>86</math>. | ||
+ | |||
+ | At <math>k=1</math>, <math>10k+93=103\;and\;86+k^2=87</math>, <math>103>87</math>. | ||
+ | |||
+ | At <math>k=2</math>, <math>10k+93=113\;and\;86+k^2=90</math>, <math>113>90</math>. | ||
+ | |||
+ | At <math>k=3</math>, <math>10k+93=123\;and\;86+k^2=95</math>, <math>123>95</math>. | ||
+ | |||
+ | At <math>k=4</math>, <math>10k+93=133\;and\;86+k^2=102</math>, <math>133>102</math>. | ||
+ | |||
+ | At <math>k=5</math>, <math>10k+93=143\;and\;86+k^2=111</math>, <math>143>111</math>. | ||
+ | |||
+ | At <math>k=6</math>, <math>10k+93=153\;and\;86+k^2=122</math>, <math>153>122</math>. | ||
+ | |||
+ | At <math>k=7</math>, <math>10k+93=163\;and\;86+k^2=135</math>, <math>163>135</math>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 with <math>0 \le a \le 9</math> and <math>0 \le b \le 9</math> | ||
+ | |||
+ | <math>n=2000+10a+b</math>; <math>S(n)=4+a^2+b^2</math> | ||
+ | |||
+ | <math>n-2007=10a+b-7 \le S(n)=4+a^2+b^2</math> | ||
+ | |||
+ | Solving the inequality <math>10a+b-7 \le 4+a^2+b^2</math> we have: | ||
+ | |||
+ | <math>0 \le b^2-b+(a^2-10a+11)</math> | ||
+ | |||
+ | Substituting for all values of a in the above inequality we get: | ||
+ | |||
+ | When <math>a=0,\;\;0 \le b^2-b+11</math>, which gives: <math>0 \le b \le 9</math>. But <math>n>2007</math>, So, <math>n=2008</math> and <math>n=2009</math>. Total possible <math>n</math>'s: '''2''' | ||
+ | |||
+ | When <math>a=1,\;\;0 \le b^2-b+2</math>, which gives: <math>0 \le b \le 9</math>. Total possible <math>n</math>'s: '''10''' | ||
+ | |||
+ | When <math>a=2,\;\;0 \le b^2-b-5</math>, which gives: <math>3 \le b \le 9</math>. Total possible <math>n</math>'s: '''7''' | ||
+ | |||
+ | When <math>a=3,\;\;0 \le b^2-b-10</math>, which gives: <math>4 \le b \le 9</math>. Total possible <math>n</math>'s: '''6''' | ||
+ | |||
+ | When <math>a=4,\;\;0 \le b^2-b-13</math>, which gives: <math>5 \le b \le 9</math>. Total possible <math>n</math>'s: '''5''' | ||
+ | |||
+ | When <math>a=5,\;\;0 \le b^2-b-14</math>, which gives: <math>5 \le b \le 9</math>. Total possible <math>n</math>'s: '''5''' | ||
+ | |||
+ | When <math>a=6,\;\;0 \le b^2-b-13</math>, which gives: <math>5 \le b \le 9</math>. Total possible <math>n</math>'s: '''5''' | ||
+ | |||
+ | When <math>a=7,\;\;0 \le b^2-b-10</math>, which gives: <math>4 \le b \le 9</math>. Total possible <math>n</math>'s: '''6''' | ||
+ | |||
+ | When <math>a=8,\;\;0 \le b^2-b-5</math>, which gives: <math>3 \le b \le 9</math>. Total possible <math>n</math>'s: '''7''' | ||
+ | |||
+ | When <math>a=9,\;\;0 \le b^2-b+2</math>, which gives: <math>0 \le b \le 9</math>. Total possible <math>n</math>'s: '''10''' | ||
+ | |||
+ | Therefore, the total number of possible <math>n</math>'s is: <math>2+10+7+6+5+5+5+6+7+10=\boxed{63}</math> | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
+ | |||
+ | {{alternate solutions}} |
Latest revision as of 23:02, 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.
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 with
and
;
Solving the inequality we have:
Substituting for all values of a in the above inequality we get:
When , which gives:
. But
, So,
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
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.