1999 AMC 12 Problem 4

Attempt Problem 4 of the 1999 AMC 12 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 1999 AMC 12 solutions, or check the answer key.

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

Find the sum of all prime numbers between

1

and

100

that are simultaneously

1

greater than a multiple of

4

and

1

less than a multiple of

5.

118

137

158

187

245

A number

1

less than a multiple of

5

has units digit

4

9

Being also

\equiv 1 \pmod 4

forces the number to be

\equiv 9 \pmod{20}

A number that is

1

less than a multiple of

5

ends in

4

9,

and one that is

1

greater than a multiple of

4

is odd. Together these give numbers

\equiv 9 \pmod{20},

namely

9, 29, 49, 69, 89.

Among these, only

29

and

89

are prime, and their sum is

29 + 89 = 118.

Thus, the correct answer is A.

Problem 4 in other years