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# I can't figure this out

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A) If $a\equiv 62\pmod{99}$ and $b\equiv 75\pmod{99}$, then for what integer $n$ in the set $\{1000,1001,1002,\ldots,1097,1098\}$ is it true that $$a-b\equiv n\pmod{99}~?$$

I have edited so that it can be read. Next time edit it yourself.    Melody.

(It still doesn't make sense though)

$$\color{blue} {If\;\;a\equiv 62\pmod{99} \;\;and \;\; b\equiv 75\pmod{99}\\ \text{ then for what integer n in the set } \{1000,1001,1002,\ldots,1097,1098\}\\\text{ is it true that }a-b\equiv n\pmod{99}~?}$$

B) The remainders when three positive integers are divided by 5 are 1, 2, and 3. Find the remainder when their product is divided by 5.

C) What is the residue modulo  of the sum of the modulo  inverses of the first positive integers?

D) Solve the congruence , as a residue modulo 43. (Give an answer between 0 and 42.)

off-topic
Dec 8, 2018
edited by Guest  Dec 8, 2018
edited by Melody  Dec 8, 2018

#1
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a) Assume a = 62 and b = 75. a - b = -13, which equals 86 (mod 99).

The problem is asking n $$\cong$$ 86 (mod 99) for the number $$n$$ in the set {1000, 1001....1098}.

We can clearly see that 990 (mod 99) = 0. So 990 + 99 = 1089, which is also divisible by 99. So if n = 1089, then $$n \cong 0 (mod99)$$. But we need $$n \cong 86 (mod99)$$, so n = 1089 - 13 = 1076. The anwer is $$\boxed{1076}$$.

Hope this helps,

- PM

edited by PartialMathematician  Dec 9, 2018
#2
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it should make sense now...

#3
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b) Just plug in numbers 1, 2, and 3.

1*2*3 = 6. 6/5 = 1 R1.

Remainder $$\boxed{1}$$.

- PM

#4
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If you want an actual explanation, feel free to ask for one.

#5
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I think you messed up on the LaTeX in C and D. I cant read it...

sorry,

- Pm