Solving Modulus Math Problems with Negative Numbers | Step-by-Step Guide

  • Thread starter Thread starter Fairy111
  • Start date Start date
  • Tags Tags
    Modulus
Fairy111
Messages
72
Reaction score
0

Homework Statement



Im confused as to how you obtain that,

77 is congruent to -1 mod 26
-77 is congruent to 1 mod 26
-11 is congruent to 15 mod 26

Homework Equations





The Attempt at a Solution



Some help would be great thanks
 
Physics news on Phys.org


haven't worked a heap with mod, but i think the trick is that the remainder is positive, so for the first
77 = 2x26 + 25, so 77mod26 = 25

simlarly
-1 = (-1)*26 + 25, then (-1)mod26 = 25
 
Last edited:


a is congruent to b mod 26 if a-b is divisible by 26. 77-(-1)=78, -77-1=(-78), -11-15=(-26). All of the differences are evenly divisible by 26.
 


to clrify, i think mod is defined by the fact that if
a mob n = b
then (b-a) is a multiple of n

so take a poitive number c, such that c<n, clearly
c mod n = c
now consider the negative case, say
(-c) mod n = b
then as c<n, and b>0, we must have
(-c)-b = (-1)n
so
b = n -c
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top