2007 AMC 12A Problem 22

For each positive integer

n,

let

S(n)

denote the sum of the digits of

n.

For how many values of

n

n+S(n)+S(S(n))=2007?

Show solution

Solution

For

n\le 2007,

S(n)\le S(1999)=28,

and then

S(S(n))\le S(28)=10.

So any solution has

n\ge 2007-28-10=1969.

Also

n,

S(n),

and

S(S(n))

are congruent modulo

9,

and

2007

is a multiple of

9,

so all three must be multiples of

3.

Checking the multiples of

3

between

1969

and

2007

(many are eliminated because

n+S(n)

already exceeds

2007

) leaves

1977,1980,1983,

and

2001.

That is

4

values. Thus, the correct answer is D.

Tagged: digits · modular arithmetic · bounding to limit cases