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# Uh, LCD and GCD

-1
183
4

(deleted)

Apr 24, 2020
edited by somehelpplease  May 8, 2020

#1
+2

a=1;c=0; d=gcd(a, 15);if(d==3, goto4, goto6);c++;printc,"-","GCD of",a,"and", "15","=",d;a++;if(a<100, goto2, 0)

OUTPUT:
1 - GCD of 3 and 15 = 3
2 - GCD of 6 and 15 = 3
3 - GCD of 9 and 15 = 3
4 - GCD of 12 and 15 = 3
5 - GCD of 18 and 15 = 3
6 - GCD of 21 and 15 = 3
7 - GCD of 24 and 15 = 3
8 - GCD of 27 and 15 = 3
9 - GCD of 33 and 15 = 3
10 - GCD of 36 and 15 = 3
11 - GCD of 39 and 15 = 3
12 - GCD of 42 and 15 = 3
13 - GCD of 48 and 15 = 3
14 - GCD of 51 and 15 = 3
15 - GCD of 54 and 15 = 3
16 - GCD of 57 and 15 = 3
17 - GCD of 63 and 15 = 3
18 - GCD of 66 and 15 = 3
19 - GCD of 69 and 15 = 3
20 - GCD of 72 and 15 = 3
21 - GCD of 78 and 15 = 3
22 - GCD of 81 and 15 = 3
23 - GCD of 84 and 15 = 3
24 - GCD of 87 and 15 = 3
25 - GCD of 93 and 15 = 3
26 - GCD of 96 and 15 = 3
27 - GCD of 99 and 15 = 3

Apr 24, 2020
#2
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Thank you!

#3
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For how many integers n between 1 and 100 is the greatest common divisor of 15 and n equal to 3?

15=3*5

So all the numbers we want are 3*N where N is not a multiple of 5

100/3 =33 and a bit

so there are 33 multiples of 3

there are 6 multiples of 15 we do not want those.

33-6 = 27

Apr 25, 2020
#4
+1

Thanks Melody! I really appreciate that you worked it out. Have a good day :D