2003 AMC 8 Problem 19

Attempt Problem 19 of the 2003 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

How many integers between 10001000 and 20002000 have all three of the numbers 15,15, 20,20, and 2525 as factors?
11
22
33
44
55
Answer: C
Concepts:least common multiplecounting integers in a range
Small hint:
Use the least common multiple of 15,15, 20,20, and 25.25.
Big hint:
Count the multiples of that LCM between 10001000 and 2000.2000.
Solution:
If a number xx has these three numbers as factors, then their least common multiple must also divide x.x.
These numbers have the following prime factorizations: 15=3⋅5, 15 = 3 \cdot 5,20=22⋅5, 20 = 2^2 \cdot 5,25=52. 25 = 5^2.
From these values, we get that the least common multiple is 22⋅3⋅52=300. 2^2 \cdot 3 \cdot 5^2 = 300.
Therefore, the multiples of 300300 between 10001000 and 20002000 are 1200,1200, 1500,1500, and 1800.1800.
Thus, C is the correct answer.
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