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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 5"
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<math>n-2007=10a+b-7 \le 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:
 +
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<math>0 \le b^2-b+(a^2-10a+11)</math>
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Revision as of 15:57, 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 $2^2+k^2 \le S(n) \le 2^2+k^2+2 \times 9^2$

$(k-1)100+93 \le n-2007 \le (k-1)100+92$, and $4+k^2 \le S(n) \le 166+k^2$

At $k=2$, $100(k-1)+93=193>166+k^2>170$.

At $k=3$, $100(k-1)+93=293>166+k^2>175$.

At $k=4$, $100(k-1)+93=393>166+k^2>182$.

At $k=5$, $100(k-1)+93=493>166+k^2>191$.

At $k=6$, $100(k-1)+93=593>166+k^2>202$.

At $k=7$, $100(k-1)+93=693>166+k^2>215$.

At $k=8$, $100(k-1)+93=793>166+k^2>230$.

At $k=9$, $100(k-1)+93=893>166+k^2>247$.

Since $100(k-1)+93 > 166+k^2$, 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.

NOTE... case 4 is wrong. Need to rewrite it

Case 4: $2110 \le n \le 2199$

Let $1 \le k \le 9$ be the 3rd digit of $n$

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

$(k-1)10+93 \le n-2007 \le (k-1)10+102$, and $5+k^2 \le S(n) \le 86+k^2$

At $k=1$, $10(k-1)+93=93>86+k^2>87$.

At $k=2$, $10(k-1)+93=103>86+k^2>90$.

At $k=3$, $10(k-1)+93=113>86+k^2>95$.

At $k=4$, $10(k-1)+93=123>86+k^2>102$.

At $k=5$, $10(k-1)+93=133>86+k^2>111$.

At $k=6$, $10(k-1)+93=143>86+k^2>122$.

At $k=7$, $10(k-1)+93=153>86+k^2>135$.

At $k=8$, $10(k-1)+93=163>86+k^2>150$.

At $k=9$, $10(k-1)+93=173>86+k^2>167$.

Since $10(k-1)+93 > 85+k^2$, for $1 \le k \le 9$, then $n-2007\not\le S(n)$ and there is no possible $n$ when $n \ge 2110$ when combined with the previous cases.

Case 5: Here we need to try each case from n=2008 to n=2109

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:

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



...ongoing writing of solution...

~Tomas Diaz. orders@tomasdiaz.com

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