Difference between revisions of "2004 AMC 10B Problems/Problem 21"
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Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>. | Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{3722}</math>. | ||
===Solution 2=== | ===Solution 2=== | ||
− | Shift down the first sequence by <math>1</math> and the second by <math>9</math> so that the two sequences become <math>0,3,6,\cdots,6009</math> and <math>0,7,14,\cdots,14028</math>. The first becomes multiples of <math>3</math> and the second becomes multiples of <math>7</math>. Their intersection is the multiples of <math>21</math> up to <math>6009</math>. There are <math>\lfloor \frac{6009}{21} \rfloor</math> multiples of <math> | + | Shift down the first sequence by <math>1</math> and the second by <math>9</math> so that the two sequences become <math>0,3,6,\cdots,6009</math> and <math>0,7,14,\cdots,14028</math>. The first becomes multiples of <math>3</math> and the second becomes multiples of <math>7</math>. Their intersection is the multiples of <math>21</math> up to <math>6009</math>. There are <math>\lfloor \frac{6009}{21} \rfloor=286</math> multiples of <math>21</math>. There are <math>4008-286=\boxed{\textbf{(A)}\ 3722}</math> distinct numbers in <math>S</math>. |
== See also == | == See also == | ||
{{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}} |
Revision as of 16:30, 27 May 2013
Problem
Let ;
;
and
;
;
be two arithmetic progressions. The set
is the union of the first
terms of each sequence. How many distinct numbers are in
?
Solution
Solution 1
The two sets of terms are and
.
Now . We can compute
. We will now find
.
Consider the numbers in . We want to find out how many of them lie in
. In other words, we need to find out the number of valid values of
for which
.
The fact "" can be rewritten as "
, and
".
The first condition gives , the second one gives
.
Thus the good values of are
, and their count is
.
Therefore , and thus
.
Solution 2
Shift down the first sequence by and the second by
so that the two sequences become
and
. The first becomes multiples of
and the second becomes multiples of
. Their intersection is the multiples of
up to
. There are
multiples of
. There are
distinct numbers in
.
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |