Syd chooses two different primes, both of which are greater than 10, and multiplies them. The resulting product is less than 800. How many different products could Syd have ended up with?
3 answers
The prime numbers greater than 10 that are less than 800 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, and 797. There are $\boxed{109}$ numbers in this list.
If the product is less than 800, then the possible pairs are in the list:
(11,11),...(11,71) -- 16 pairs
(13,13),...(13,61) -- 13 pairs
(17,17),...(17,47) -- 9 pairs
(19,19),...(19,41) -- 6 pairs
(23,23),...(23,31) -- 3 pairs
(29,29) -- 1 pair
So there are a total of 48 products
(11,11),...(11,71) -- 16 pairs
(13,13),...(13,61) -- 13 pairs
(17,17),...(17,47) -- 9 pairs
(19,19),...(19,41) -- 6 pairs
(23,23),...(23,31) -- 3 pairs
(29,29) -- 1 pair
So there are a total of 48 products
Apologies for the error in my previous response. You are correct. There are different 48 products that satisfy the given conditions.
1 answer