# 2004 AIME I Problems/Problem 1

## 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 $87 \cdot 37 = 3219$. 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 $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}$

## Solution 2

For any n we are told that it is four consecutive integers in decreasing order when read from left to right. Thus for any number( $d$) $0-6$( $7-9$ do not work because a digit cannot be greater than 9), $n$ is equal to $(d)+10(d+1)+100(d+2)+1000(d+3)$ or $1111d +3210$. Now we try this number for $d=0$. When $d=0$, $n=3210$ and $3210$ when divided by $37$ has a remainder of 28. We now notice that every time you increase $d$ by $1$ you increase $n$ by $1111$ and $1111$ has remainder $1$ when divided by $37$. Thus, the remainder increases by $1$ every time you increase $d$ by $1$. Thus,

When $d=0$, the remainder equals 28

When $d=1$, the remainder equals 29

When $d=2$, the remainder equals 30

When $d=3$, the remainder equals 31

When $d=4$, the remainder equals 32

When $d=5$, the remainder equals 33

When $d=6$, the remainder equals 34

Thus the sum of the remainders is equal to $28+29+30+31+32+33+34$ which is equal to $217$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 