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

Jasonkiln Aug 2, 2022

#1**+1 **

(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**

Guest Aug 2, 2022

edited by
Guest
Aug 2, 2022

#2**0 **

Thanks for listing them out but do you know any way to do it without listing them out? Because that could take a while

Jasonkiln Aug 2, 2022

#5**+1 **

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.

BuilderBoi Aug 3, 2022