Mods.

**Showcase_22** · April 28th 2009, 02:23 AM

FInd an integer m satisfying $n \leq m \leq n+11$ and $m \equiv 7(mod \ 12)$ in each of the following questions.

(i) $n=100$

$10=(-2) \ mod \ 12$ where $-n<r \leq 0$ for $x \equiv r (mod \ n)$ .

$\Rightarrow 10^2=4 \ mod \ 12$ so I need the number to be $100+3=103.$

(ii). $n=10^6$

$10=-2 \ mod \ 12$ where $-n<r \leq 0$ for $x \equiv r (mod \ n)$ .

$\Rightarrow \ 10^6=(-2)^6 \ mod \ 12 \Rightarrow 10^6=64 \ mod \ 12 \Rightarrow \ 10^6=4 \ mod \ 12$ .

But I require a number m such that $m=7 \ mod \ 12.$

Combining my above results reveals the number I need to be $10^6+3$ .

$(iii). n=-999$

$-9= 3 \ mod \ 12 \Rightarrow \ -9(111)=3(111) \ mod \ 12 \Rightarrow \ -999=9 \ mod \ 12$

Comparing this with $7 \ mod \ 12$ gives $m=-999-2=-1001$ .

My method seems to be working pretty well (i've checked them all on a calculator) but is there an easier way to do them?

**HallsofIvy** · April 28th 2009, 02:43 AM

Strictly speaking, if the problem is just "find a number, m, such that 0< m< n+ 11 and m= 7 (mod 12)" then "m= 7" is a perfectly valid answer for all n! Did you mean "such that n< m< n+ 11"?

Any number m, satisfying m= 7 (mod 12) can be written m= 7+ 12k for some integer k. To find such a number such that n< m< n+ 11, divide n by 12 to see how large k must be. For example, with $n= 10^6$ , the largest value of n that you have here, 12 divides into [tex]10^6= 1000000[/itex] 83333 times. Taking k= 83333, we have 7+ 12(83333)= 100003.

For n= -999, divide -999 by 12: 12 divides into -999 -83 times. Taking k= -83, we have 7+ 12(-83)= -989 which satisfies -999< -989< -988.

**Showcase_22** · April 28th 2009, 02:47 AM

Ah yes I did sorry!!

One of the problems is that i'm not supposed to use a calculator, I did all the ones in my previous post using mental artihmetic.

I'm just pretty worried that when I get tested on this I might have a question where instead of $10^6$ or $10^2$ , I get a number like $10^{20}$ which is going to be a pain to do!

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