1999 AHSME Problems/Problem 4
Problem
Find the sum of all prime numbers between 
and 
that are simultaneously greater than a multiple of 
and less than a multiple of 
.
Solution
Numbers that are less than a multiple of all end in or 
.
No prime number ends in , since all numbers that end in are divisible by 
. Thus, we are only looking for numbers that end in .
Writing down the ten numbers that so far qualify, we get 
.
Crossing off multiples of 
gives 
.
Crossing off numbers that are not more than a multiple of (in other words, numbers that are less than a multiple of , since all numbers are odd), we get:
Noting that 
is not prime, we have only 
and 
, which give a sum of 
, so the answer is 
.
See Also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 5 | |
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