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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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Substituting for all values of a in the above inequality we get:
 
Substituting for all values of a in the above inequality we get:
  
−
When <math>a=0,\;</math>, <math>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=0,\;</math>, <math>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,\;</math>, <math>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=1,\;</math>, <math>0 \le b^2-b+2</math>, which gives: <math>0 \le b \le 9</math>.  Total possible <math>n</math>'s: '''10'''

Revision as of 21:57, 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 with $0 \le a \le 9$ and $0 \le b \le 9$

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

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

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

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

Substituting for all values of a in the above inequality we get:

When $a=0,\;$, $0 \le b^2-b+11$, which gives: $0 \le b \le 9$. But $n>2007$, So, $n=2008$ and $n=2009$. Total possible $n$'s: 2

When $a=1,\;$, $0 \le b^2-b+2$, which gives: $0 \le b \le 9$. Total possible $n$'s: 10

When $a=2,\;$, $0 \le b^2-b-5$, which gives: $3 \le b \le 9$. Total possible $n$'s: 7

When $a=3,\;$, $0 \le b^2-b-10$, which gives: $4 \le b \le 9$. Total possible $n$'s: 6

When $a=4,\;$, $0 \le b^2-b-13$, which gives: $5 \le b \le 9$. Total possible $n$'s: 5

When $a=5,\;$, $0 \le b^2-b-14$, which gives: $5 \le b \le 9$. Total possible $n$'s: 5

When $a=6,\;$, $0 \le b^2-b-13$, which gives: $5 \le b \le 9$. Total possible $n$'s: 5

When $a=7,\;$, $0 \le b^2-b-10$, which gives: $4 \le b \le 9$. Total possible $n$'s: 6

When $a=8,\;$, $0 \le b^2-b-5$, which gives: $3 \le b \le 9$. Total possible $n$'s: 7

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

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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