Mock AIME 6 2006-2007 Problems/Problem 5

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.

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:

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

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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Mock AIME 6 2006-2007 Problems/Problem 5

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