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: $2200 \le n \le 2299$

Let $2 \le k \le 9$ be the 3rd digit of $n$

$2000+10k \le n \le 2009+10k$ , and $2^2+k^2 \le S(n) \le 2^2+k^2+9^2$

$(k-1)10+3 \le n-2007 \le (k-1)10+2$ , and $4+k^2 \le S(n) \le 85+k^2$

At $k=2$ , $10(k-1)+3=193>85+k^2>89$ .

At $k=3$ , $10(k-1)+3=293>85+k^2>94$ .

At $k=4$ , $10(k-1)+3=393>85+k^2>101$ .

At $k=5$ , $10(k-1)+3=493>85+k^2>110$ .

At $k=6$ , $10(k-1)+3=593>85+k^2>121$ .

At $k=7$ , $10(k-1)+3=693>85+k^2>134$ .

At $k=8$ , $10(k-1)+3=793>85+k^2>149$ .

At $k=9$ , $10(k-1)+3=893>85+k^2>166$ .

Since $10(k-1)+3 > 85+k^2$ , for $2 \le k \le 9$ , then $n-2007\not\le S(n)$ and there is no possible $n$ when $n \ge 2120$ when combined with the previous cases.

...ongoing writing of solution...

~Tomas Diaz. orders@tomasdiaz.com

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Mock AIME 6 2006-2007 Problems/Problem 5

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