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Determine the lowest numbet n>5 so that n≡5 (mod 36). Also explain how and why thank you

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Determine the lowest numbet n>5 so that n≡5 (mod 36). Also explain how and why thank you

Mar 17, 2015

#1
+17774
+10

When you are looking for the answer, mode 36, you are looking for a number whose remainder is 5 when that number is divided by 36.

All the numbers from 0 through 35, when divided by 36, have 0 for a quotient and that number for a remainder.

36 mod 36 is 0 because 36 divided by 36 is 1 with a remainder of 0.

37 mod 36 is 1 because 37 divided by 36 is 1 with a remainder of 1.

Continuing:

38 mod 36 is 2 because 38 divided by 36 is 1 with a remainder of 2.

39 mod 36 is 3 because 39 divided by 36 is 1 with a remainder of 3.

40 mod 36 is 4 because 40 divided by 36 is 1 with a remainder of 4.

41 mod 36 is 1 becauae 41 divided by 36 is 1 with a remainder of 5.  (Finally!)

-- I could have gotten there quicker just by adding 5 to 36!

Mar 17, 2015

#1
+17774
+10

When you are looking for the answer, mode 36, you are looking for a number whose remainder is 5 when that number is divided by 36.

All the numbers from 0 through 35, when divided by 36, have 0 for a quotient and that number for a remainder.

36 mod 36 is 0 because 36 divided by 36 is 1 with a remainder of 0.

37 mod 36 is 1 because 37 divided by 36 is 1 with a remainder of 1.

Continuing:

38 mod 36 is 2 because 38 divided by 36 is 1 with a remainder of 2.

39 mod 36 is 3 because 39 divided by 36 is 1 with a remainder of 3.

40 mod 36 is 4 because 40 divided by 36 is 1 with a remainder of 4.

41 mod 36 is 1 becauae 41 divided by 36 is 1 with a remainder of 5.  (Finally!)

-- I could have gotten there quicker just by adding 5 to 36!

geno3141 Mar 17, 2015
#2
+5

Allright, i get it now! Thanks a ton!

Mar 17, 2015
#3
+101741
0

Thanks Geno, that is a really good explanation.     ヽ(ヅ)ノ

Mar 18, 2015