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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
Line 63: Line 63:
  
  
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'''Case 4:''' <math>2110 \le n \le 2199</math>
+
'''Case 4:''' <math>2100 \le n \le 2199</math>
  
−
Let <math>1 \le k \le 9</math> be the 3rd digit of <math>n</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+k^2 \le S(n) \le 2^2+1^2+k^2+9^2</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-7 \le n-2007 \le 10l+102</math>, and <math>5+k^2 \le S(n) \le 86+k^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=1</math>, <math>10(k-1)+93=93>86+k^2>87</math>.
+
At <math>k=0</math>, <math>10k+93=93\;and\;86+k^2=86</math>, <math>93>86</math>.
  
−
At <math>k=2</math>, <math>10(k-1)+93=103>86+k^2>90</math>.
+
At <math>k=1</math>, <math>10k+93=103\;and\;86+k^2=87</math>, <math>103>87</math>.
  
−
At <math>k=3</math>, <math>10(k-1)+93=113>86+k^2>95</math>.
+
At <math>k=2</math>, <math>10k+93=113\;and\;86+k^2=90</math>, <math>113>90</math>.
  
−
At <math>k=4</math>, <math>10(k-1)+93=123>86+k^2>102</math>.
+
At <math>k=3</math>, <math>10k+93=123\;and\;86+k^2=95</math>, <math>123>95</math>.
  
−
At <math>k=5</math>, <math>10(k-1)+93=133>86+k^2>111</math>.
+
At <math>k=4</math>, <math>10k+93=133\;and\;86+k^2=102</math>, <math>133>102</math>.
  
−
At <math>k=6</math>, <math>10(k-1)+93=143>86+k^2>122</math>.
+
At <math>k=5</math>, <math>10k+93=143\;and\;86+k^2=111</math>, <math>143>111</math>.
  
−
At <math>k=7</math>, <math>10(k-1)+93=153>86+k^2>135</math>.
+
At <math>k=6</math>, <math>10k+93=153\;and\;86+k^2=122</math>, <math>153>122</math>.
  
−
At <math>k=8</math>, <math>10(k-1)+93=163>86+k^2>150</math>.
+
At <math>k=7</math>, <math>10k+93=163\;and\;86+k^2=135</math>, <math>163>135</math>.
  
−
At <math>k=9</math>, <math>10(k-1)+93=173>86+k^2>167</math>.
+
At <math>k=8</math>, <math>10k+93=173\;and\;86+k^2=150</math>, <math>173>150</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.
+
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.
  
−
'''Case 5:'''
+
From cases 1 through 4 we now know that <math>2008 \le n \le 2099</math>
−
Here we need to try each case from n=2008 to n=2109
+
 
 +
'''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.
 
Let <math>a</math> and <math>b</math> be the 3rd and 4th digits of n respectively.

Revision as of 22:47, 24 November 2023

Problem

Let $S(n)$ be the sum of the squares of the digits of $n$. How many positive integers $n>2007$ satisfy the inequality $n-S(n)\le 2007$?

Solution

We start by rearranging the inequality the following way:

$n-2007\le S(n)$ and compare the possible values for the left hand side and the right hand side of this inequality.


Case 1: $n$ has 5 digits or more.

Let $d$ = number of digits of n.

Then as a function of d,

$10^d \le n < 10^{d+1}-1$, and $1 \le S(n) \le 9^2d$

$10^d - 2007 \le n-2007 < 10^{d+1}-2008$, and $1 \le S(n) \le 81d$

when $d \ge 5$,

$10^d - 2007 \ge 10^5 -2007$

$10^d - 2007 \ge 10^5 -2007 > 81d$

Since $10^d - 2007 > 81d$ for $d \ge 5$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n$ has 5 or more digits.


Case 2: $n$ has 4 digits and $n \ge 3000$

$3000 \le n \le 9999$, and $3^2 \le S(n) \le 3^2+3 \times 9^2$

$993 \le n-2007 \le 7992$, and $9 \le S(n) \le 252$

Since $993 > 252$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n$ has 4 digits and $n \ge 3000$.


Case 3: $2200 \le n \le 2999$

Let $2 \le k \le 9$ be the 2nd digit of $n$

$2000+100k \le n \le 2099+100k$, and $2^2+k^2 \le S(n) \le 2^2+k^2+2 \times 9^2$

$100k-7 \le n-2007 \le 100k+92$, and $4+k^2 \le S(n) \le 166+k^2$

At $k=2$, $100k-7=193\;\;86+k^2=170$, $193>170$.

At $k=3$, $100k-7=293\;\;86+k^2=175$, $293>175$.

At $k=4$, $100k-7=393\;\;86+k^2=182$, $393>182$.

At $k=5$, $100k-7=493\;\;86+k^2=191$, $493>191$.

At $k=6$, $100k-7=593\;\;86+k^2=202$, $593>202$.

At $k=7$, $100k-7=693\;\;86+k^2=215$, $693>215$.

At $k=8$, $100k-7=793\;\;86+k^2=230$, $793>230$.

At $k=9$, $100k-7=893\;\;86+k^2=247$, $893>247$.

Since $100k-7 > 166+k^2$, for $2 \le k \le 9$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n \ge 2200$ when combined with the previous cases.


Case 4: $2100 \le n \le 2199$

Let $0 \le k \le 9$ be the 3rd digit of $n$

$2100+10k \le n \le 2109+10k$, and $2^2+1^2+k^2 \le S(n) \le 2^2+1^2+k^2+9^2$

$10k+93 \le n-2007 \le 10k+102$, and $5+k^2 \le S(n) \le 86+k^2$

At $k=0$, $10k+93=93\;and\;86+k^2=86$, $93>86$.

At $k=1$, $10k+93=103\;and\;86+k^2=87$, $103>87$.

At $k=2$, $10k+93=113\;and\;86+k^2=90$, $113>90$.

At $k=3$, $10k+93=123\;and\;86+k^2=95$, $123>95$.

At $k=4$, $10k+93=133\;and\;86+k^2=102$, $133>102$.

At $k=5$, $10k+93=143\;and\;86+k^2=111$, $143>111$.

At $k=6$, $10k+93=153\;and\;86+k^2=122$, $153>122$.

At $k=7$, $10k+93=163\;and\;86+k^2=135$, $163>135$.

At $k=8$, $10k+93=173\;and\;86+k^2=150$, $173>150$.

At $k=9$, $10k+93=183\;and\;86+k^2=167$, $183>167$.

Since $10k+93 > 85+k^2$, for $0 \le k \le 9$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n \ge 2100$ when combined with the previous cases.

From cases 1 through 4 we now know that $2008 \le n \le 2099$

Case 5: $2008 \le n \le 2099$

Let $a$ and $b$ be the 3rd and 4th digits of n respectively.

$n=2000+10a+b$; $S(n)=4+a^2+b^2$

$n-2007=10a+b-7 \le S(n)=4+a^2+b^2$

Solving the inequality $10a+b-7 \le 4+a^2+b^2$ we have:

$0 \le b^2-b+(a^2-10a+11)$

When $a=0\;$, $0 \le b^2-b+11$, which gives: $b \ge 0$. Which is $n=2008$ and $n=2009$ Total possible $n$'s: 2

When $a=1$, $0 \le b^2-b+2$, which gives: $b \ge 0$. Total possible $n$'s: 10

When $a=2$, $0 \le b^2-b-5$, which gives: $b \ge 3$. Total possible $n$'s: 7

When $a=3$, $0 \le b^2-b-10$, which gives: $b \ge 4$. Total possible $n$'s: 6

When $a=4$, $0 \le b^2-b-13$, which gives: $b \ge 5$. Total possible $n$'s: 5

When $a=5$, $0 \le b^2-b-14$, which gives: $b \ge 5$. Total possible $n$'s: 5

When $a=6$, $0 \le b^2-b-13$, which gives: $b \ge 5$. Total possible $n$'s: 5

When $a=7$, $0 \le b^2-b-10$, which gives: $b \ge 4$. Total possible $n$'s: 6

When $a=8$, $0 \le b^2-b-5$, which gives: $b \ge 3$. Total possible $n$'s: 7

When $a=9$, $0 \le b^2-b+2$, which gives: $b \ge 0$. Total possible $n$'s: 10

No valid $n$ for $2100 \le n \le 2109$

Therefore, the total number of possible $n$'s is: $2+10+7+6+5+5+5+6+7+10=\boxed{63}$

~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.

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