+275 How many three-digit multiples of 5 have three different digits and at least one prime digit?
Contuing on from last time. there are 720 numbers that have atleast one prime digit **the 720 is not three different digits** The part that I am stuck with is three digit multiples of 5. I thought about listing them
(105, 120, 125, 130, 135, 145, 150, 165, 170, 175, 185, 195, 205, 210, 215, 230, 235, 240, 245, 250, 260, 265, 270, 275, 280, 285, 290, 295, 305, 310, 315, 320, 325, 340, 345, 350, 360, 365, 370, 375, 380, 385, 390, 395, 405, 415, 420, 425, 430, 435, 450, 465, 470, 475, 485, 495, 510, 520, 530, 540, 560, 570, 580, 590, 605, 615, 620, 625, 630, 635, 645, 650, 670, 675, 685, 695, 705, 710, 715, 720, 725, 730, 735, 740, 745, 750, 760, 765, 780, 785, 790, 795, 805, 815, 820, 825, 830, 835, 845, 850, 865, 870, 875, 895, 905, 915, 920, 925, 930, 935, 945, 950, 965, 970, 975, 985)==116 such 3-digit integers
+275 Thanks for listing them out but do you know any way to do it without listing them out? Because that could take a while
+2668 Start with numbers ending in 5.
The prime restriction is already satisfied, so we don't worry about that.
Then, there are 8 choices for the hundreds digit (anything but 0 or 5) and 8 for the tens digit (anything but the hundreds digit and 5), which makes for \(8 \times 8 = 64\)
Now, if the final digit is 0, there are 9 choices for the hundreds (anything but 0), and 8 for the tens (anything but the hundreds and 0), which makes for \(9 \times 8 = 72\)
But, we need to account for numbers that have no primes. The digits that aren't prime are (0, 1, 4, 6, 8, and 9).
There are 5 choices for the hundreds and 4 for the tens, which makes for \(5 \times 4 = 20\) numbers that don't work.
So, there are \(64 + 72 - 20 = \color{brown}\boxed{116}\) numbers that work.
