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### Princeton Review Error : Congruences Page 225

Posted: **Mon Sep 21, 2009 3:05 am**

by **Legendre**

**Princeton Review, rule 3 :**

For any positive integer c, the statement

'a is congruent to b (mod n)'

<=>

'a is congruent to b (mod cn)'

** Counter-example : **

5 congruent 2 (mod 3)

Let c = 2.

5 congruent 2 (mod 9) is FALSE. Since (5-2) = 3 is not divisible by 9.

### Re: Princeton Review Error : Congruences Page 225

Posted: **Tue Sep 22, 2009 11:51 pm**

by **hopeful_statguy**

What edition do you have? In the 3rd edition, rule 3 is:

If $$a_1 \equiv b_1 \pmod{n}$$ and $$a_2 \equiv b_2 \pmod{n}$$ , then

$$a_1 \pm a_2 \equiv b_1 \pm b_2 \pmod{n}$$

$$a_1 a_2 \equiv b_1 b_2 \pmod{n}$$

### Re: Princeton Review Error : Congruences Page 225

Posted: **Thu Oct 29, 2009 4:52 pm**

by **kosuke**

Sorry to bring this post back from the dead, but I'm having trouble figuring out what LeDuc meant to say here... (I think Legendre meant to say rule #4, not 3 on page 225).

Any idea on what LeDuc really means?

### Re: Princeton Review Error : Congruences Page 225

Posted: **Thu Oct 29, 2009 5:05 pm**

by **kosuke**

Sorry, nevermind.

I figured out what I was doing wrong.

### Re: Princeton Review Error : Congruences Page 225

Posted: **Fri Aug 24, 2012 6:31 am**

by **Legendre**

Sorry I meant **rule 4**, not 3.

I think he meant that a = b (mod n) then, a = b **or** b + n **or** b + 2n **or **... **or** b + (c-1)n (mod cn).

This is clear because for b + cn, b + (c+1)n,... we can mod out the cn and obtain same expression in the list.