2004 AIME I Problems/Problem 1

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Problem

The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$ ?

Solution

A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace$ .

Now, note that $3\cdot 37 = 111$ so $30 \cdot 37 = 1110$ , and $90 \cdot 37 = 3330$ so $89 \cdot 37 = 3219$ . So the remainders are all congruent to $n - 9 \pmod{37}$ . However, these numbers are negative for our choices of $n$ , so in fact the remainders must equal $n + 28$ .

Adding these numbers up, we get $0 + 1 + 2 + 3 + 4 + 5 + 6 + 7\cdot28 = \boxed{217}$ , our answer.

2004 AIME I (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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2004 AIME I Problems/Problem 1

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Problem

Solution

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