Difference between revisions of "2004 AIME I Problems/Problem 1"
(→Solution) |
m (→Solution) |
||
(10 intermediate revisions by 2 users not shown) | |||
Line 7: | Line 7: | ||
Now, note that <math>3\cdot 37 = 111</math> so <math>30 \cdot 37 = 1110</math>, and <math>90 \cdot 37 = 3330</math> so <math>87 \cdot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <math>n</math>, so in fact the remainders must equal <math>n + 28</math>. | Now, note that <math>3\cdot 37 = 111</math> so <math>30 \cdot 37 = 1110</math>, and <math>90 \cdot 37 = 3330</math> so <math>87 \cdot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <math>n</math>, so in fact the remainders must equal <math>n + 28</math>. | ||
− | Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>, | + | Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math> |
+ | |||
+ | ==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(<math>d</math>) <math>0-6</math>(<math>7-9</math> do not work because a digit cannot be greater than 9), <math>n</math> is equal to <math>(d)+10(d+1)+100(d+2)+1000(d+3)</math> or <math>1111d +3210</math>. Now we try this number for <math>d=0</math>. When <math>d=0</math>, <math>n=3210</math> and <math>3210</math> when divided by <math>37</math> has a remainder of 28. We now notice that every time you increase <math>d</math> by <math>1</math> you increase <math>n</math> by <math>1111</math> and <math>1111</math> has remainder <math>1</math> when divided by <math>37</math>. Thus, the remainder increases by <math>1</math> every time you increase <math>d</math> by <math>1</math>. Thus, | ||
+ | |||
+ | When <math>d=0</math>, the remainder equals 28 | ||
+ | |||
+ | When <math>d=1</math>, the remainder equals 29 | ||
+ | |||
+ | When <math>d=2</math>, the remainder equals 30 | ||
+ | |||
+ | When <math>d=3</math>, the remainder equals 31 | ||
+ | |||
+ | When <math>d=4</math>, the remainder equals 32 | ||
+ | |||
+ | When <math>d=5</math>, the remainder equals 33 | ||
+ | |||
+ | When <math>d=6</math>, the remainder equals 34 | ||
+ | |||
+ | |||
+ | 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>. | ||
== See also == | == See also == |
Latest revision as of 14:53, 27 December 2019
Contents
Problem
The digits of a positive integer are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when
is divided by
?
Solution
A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form , for
.
Now, note that so
, and
so
. So the remainders are all congruent to
. However, these numbers are negative for our choices of
, so in fact the remainders must equal
.
Adding these numbers up, we get
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()
(
do not work because a digit cannot be greater than 9),
is equal to
or
. Now we try this number for
. When
,
and
when divided by
has a remainder of 28. We now notice that every time you increase
by
you increase
by
and
has remainder
when divided by
. Thus, the remainder increases by
every time you increase
by
. Thus,
When , the remainder equals 28
When , the remainder equals 29
When , the remainder equals 30
When , the remainder equals 31
When , the remainder equals 32
When , the remainder equals 33
When , the remainder equals 34
Thus the sum of the remainders is equal to which is equal to
.
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.