The college academic club is holding a 50-50 raffle. In the raffle, the winner gets half the money collected from ticket sales. They are selling 1331 tickets numbered 1 through 1331. The winner will be determined by a roll of three specially created dice which have faces numbered 1, 2, 3, 5, 7, and 11. The numbers turning up on the three dice will be multiplied together to determine the winning ticket number. assume you are given first choice of buying tickets. What is the least number of tickets you must buy to be assured of obtaining all potential winning numbers?

Solution

The only numbers possible are of the form:
(1^a)(2^b)(3^c)(5^d)(7^e)(11^f) 
constrained by a + b + c + d + e + f = 3,  (only 3 dice)

where a, b, c, d, e, and f are non-negative. This is an ordered partition with 6 elements of 3. So there are    ₈C₃ = 56 of them. Thus there are 56 possibilities providing 56 unique numbers. You will need to purchase only 56 tickets.

1³  2³  3³ 5³ 7³ 11³
1²2 2²1 3²1 5²1 7²1 11²1
1²3 2²3 3²2 5²2 7²1 11²2
1²5 2²5 3²5 5²3 7²2 11²3
1²7 2²7 3²7 5²7 7²3 11²5
1²11 2²11 3²11 5²11 7²5 11²7
123 235 357 57,11 7²11
125 237 37,11
127 23,11
12,11 257
135 25,11
137 27,11
13, 11
157
15, 11
17, 11

         We received no (as in none!) puzzle solutions from the last issue. So to all of us mathematical types it is fairly obvious that No One is the winner of the drawing for the pizza. Congratulations!

Give the prime factorization of

Solution

4¹⁸ - 3¹⁸ =
              = (4⁹ - 3⁹)(4⁹ + 3⁹) = (4³ - 3³)(4⁶ + 4³3³ + 3⁶) (4³ + 3³)(4⁶ - 4³3³ + 3⁶)
              = (19)(37)(91)(163)(6553)