Difference between revisions of "2004 AIME I Problems/Problem 1"

Revision as of 22:19, 21 July 2018

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 $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}$ , our answer.

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 $217$ .

Solution 3(bashy)

Just divide every possible number(There are only 7) by 37 and add the resulting remainders.

@@ Line 31: / Line 31: @@
 Thus the sum of the remainders is equal to <math>28+29+30+31+32+33+34</math> which is equal to <math>217</math>.
-==Solution 3(bashy)
+==Solution 3(bashy)==
 Just divide every possible number(There are only 7) by 37 and add the resulting remainders.

2004 AIME I (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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