Difference between revisions of "2004 AMC 10B Problems/Problem 21"
Hithere22702 (talk | contribs) |
Tiankaizhang (talk | contribs) (→Problem) |
||
Line 4: | Line 4: | ||
<math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math> | <math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. | The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. |
Latest revision as of 23:37, 24 May 2021
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.
~Hithere22702
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.