Difference between revisions of "2004 AMC 10B Problems/Problem 21"
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==Problem== | ==Problem== | ||
| − | Let <math>1</math>; <math>4</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>? | + | Let <math>1</math>; <math>4</math>; <math>7</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>23</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>? |
<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> | ||
Revision as of 15:20, 5 May 2024
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 | ||
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