1) How many positive three-digit integers with a 5 in the units place are divisible by 15 ?

2) If a and b are integers such that \(ab\equiv 17\pmod{20}\), then what is the remainder when (a+10)(b+10) is divided by 20?

THANKS! :)

Guest Aug 6, 2022

#1**+1 **

On ex. 1 python says:

There are 33 integers in that group, they are:

15, 45, 75, 105, 135, 165, 195, 225, 255, 285, 315, 345,

375, 405, 435, 465, 495, 525, 555, 585, 615, 645, 675,

705, 735, 765, 795, 825, 855, 885, 915, 945, 975

See https://www.online-python.com/nxgpkMajEo

On ex. 2:

Turning to mod,

in python (and other programming languages)

% is used as the modulus operator

ab%20 = 17

(a + 10) x (b + 10)%20 =

(ab +10a + 10b + 100)%20 =

ab%20 + 10a%20 + 10b%20 + 100%20 =

17 + 10 + 10 + 0 = 17 + 20 = 37 (assuming a and b not divisible by 20)

But 37%20 = 17

tuffla2022 Aug 6, 2022

edited by
tuffla2022
Aug 6, 2022

edited by tuffla2022 Aug 6, 2022

edited by tuffla2022 Aug 6, 2022

edited by tuffla2022 Aug 6, 2022

edited by tuffla2022 Aug 6, 2022

edited by tuffla2022 Aug 6, 2022

edited by tuffla2022 Aug 6, 2022

#3**+1 **

Sorry, didn't notice that there had to be 3 digits. I stand corrected by Guest below. Code has been updated.

Exercise 1:

There are 30 integers in that group, they are:

105, 135, 165, 195, 225, 255, 285, 315, 345, 375, 405, 435,

465, 495, 525, 555, 585, 615, 645, 675, 705, 735, 765,

795, 825, 855, 885, 915, 945, 975

tuffla2022
Aug 6, 2022

#2**+1 **

1) - There are:

(105, 135, 165, 195, 225, 255, 285, 315, 345, 375, 405, 435, 465, 495, 525, 555, 585, 615, 645, 675, 705, 735, 765, 795, 825, 855, 885, 915, 945, 975) >>Total = 30 such integers.

Guest Aug 6, 2022

#4**+1 **

Using python codes is a great way to efficiently solve them but we are focused on problem solving our own minds here.

1. 15 = 5 *3

Since the three digit numbers have a unit digits of 5, we just need the number in the form of AB5, to be divisible by 3. And in order for a number to be divisible by 3, its digits need to add up to a multiple of 3.

A + B + 5 = 0 (mod 3)

A + B = 1 (mod 3)

Can you solve the question from here and find all digits A and B that make the congruence true? (Hint, casework!)

2. One simple way to solve this is by using the loss of generality, plug in any numbers a and b, where their product leaves a remainder of 17 when divided by 20. After finding those numbers, plug it into (a + 10)(b + 10) and see the remainder.

Voldemort Aug 6, 2022