Difference between revisions of "2004 AMC 10B Problems/Problem 21"
Tiankaizhang (talk | contribs) (→Problem) |
Hithere22702 (talk | contribs) m (→Solution 2) |
||
| Line 31: | Line 31: | ||
terms in total. <math>4008-286=\boxed{(A) 3722}</math> | terms in total. <math>4008-286=\boxed{(A) 3722}</math> | ||
| − | ~ | + | ~kempwood |
| + | |||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:04, 23 December 2023
Contents
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 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
We can start by finding the first non-distinct term from both sequences. We find that that number is . Now, to find every
other non-distinct terms, we can just keep adding 
. We know that the last terms of both sequences are 
and
. Clearly, is smaller and that is the last possible common term of both sequences. Now, we can
create the inequality 
. Using the inequality, we find that there are 
common terms. There are 4008
terms in total. 
~kempwood
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
