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?
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
Apologies for the error in my previous response. You are correct. There are different 48 products that satisfy the given conditions.
To find the answer, we need to determine the primes greater than 10 that are less than 800. Let's list the primes greater than 10 and less than 800:
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, 797.
Now, let's consider all the possible products that can be obtained by multiplying two different primes from this list.
We can use the combination formula to find the number of combinations of 2 primes:
nC2 = n! / ((n - r)! * r!)
Here, n is the total number of primes in the list and r is the number of primes needed to form a product.
Using this formula, we find:
nC2 = 168 / ((168 - 2)! * 2!) = 168 / (166! * 2 * 1) = 168 / (332 * 2) = 168 / 664 = 0.253
Since we can't have a fraction of a combination, we round down to the nearest whole number.
Therefore, Syd could have ended up with 0 different products.