Prove or disprove the congruence below:
$$15 + 111^5· (−10)\equiv 5 \pmod{11}$$
I am not sure where to start because we cannot use a calculator this problem? Can anyone guide me?
Asked
Active
Viewed 265 times
3 Answers
2
(Since the OP asked, I am adding more detail here that Brian M. Scott has already put in his answer. He wrote it first!)
Even without a calculator we easily see that $15=1\cdot 11+4$, so
$$15\equiv 4 \pmod {11}$$
Similarly, $111=10\cdot 11+1$, so
$$111\equiv 1 \pmod {11}$$
And $-10=-1\cdot 11+1$, so
$$-10\equiv 1 \pmod {11}$$
Substituting all those into the original expression,
$$15+111^5\cdot (-10)\equiv 4+1^5\cdot(1)\equiv 4+1\equiv 5\pmod {11}$$
Rory Daulton
- 32,288
-
so we know $15≡4(mod11)$ because $15-11=4$ and $−10≡1(mod11)$ because $-10+11=1$ how do you get $111≡1(mod11)$? – Csci319 Apr 04 '15 at 19:58
-
@Csci319, $111=11\times 10+1\equiv 0+1\equiv 1\pmod{11}$ – Prasun Biswas Apr 04 '15 at 20:00
2
Replace all arguments of sums and products by their remainders mod $11$. The remainders are
$$ \begin{align} \color{#90f}{15} &\,=\, 11\cdot 1 \color{#90f}{\,+\, 4}\,\equiv\,\color{#90f}{ 4}\\ \color{#0a0}{111} &\,=\, 11\cdot 10 \color{#0a0}{+ 1}\equiv\,\color{#0a0}1\\ \color{#c00}{-10} &\,=\, 11(-1)\color{#c00}{\!+\! 1}\equiv\,\color{#c00}1\\ \\ {\rm Thus}\ \ &\quad\ \color{#90f}{15}+\color{#0a0}{111}^5\color{#c00}{(-10)}\\ \quad\ \ &\equiv\ \color{#90f}4 +\ \color{#0a0}{(1)}^5\,\color{#c00}{(1)}\\ \quad\ \ &\equiv\ 5\end{align}\qquad $$
Such argument replacements are valid by the Congruence Sum and Product Rules, i.e.
$$\color{#0a0}{A\equiv a},\ \color{}{B\equiv b}\,\Rightarrow\, \color{#0a0}A\color{}B\equiv \color{#0a0}a\color{}b,\,\ \color{#0a0}A+\color{#}B\equiv \color{#0a0}a+\color{#}b\qquad $$
or, more generally, composition of such, i.e. the Polynomial Congruence Rule.
Bill Dubuque
- 272,048
-
-
@Csci319 Right, there $,q,$ is the quotient and $,r,$ is the remainder in the division $, a \div d.,$ When computing mod $,d,$ we need only the remainder (because the quotient is multiplied by $,d,$ so their product $,dq\equiv 0\pmod d\ $ Essentially it is an arithmetic of remainders on division by $,d.\ \ $ – Bill Dubuque Apr 04 '15 at 20:18
-
so since all the remainders equal $5$ it is true, what is the remainders equaled $10$? would it still be true, so no because we need it to equal $5$? – Csci319 Apr 04 '15 at 20:21
-
@Csci319 The whole expression leaves remainder $5$. If you change the arguments of its constituent sums and products to numbers with different remainders then the remainder of the expression will generally change too . – Bill Dubuque Apr 04 '15 at 20:25
-
so when asked to prove or disprove a problem above you need to generally just find the remainders and see if it is the same – Csci319 Apr 04 '15 at 20:26
-
@Csci319 That's usually the simplest way to do it, because the arithmetic reduces to arithmetic on much smaller numbers. Note that both sides of the congruence need to be reduced to the canonical remainder interval $\ 0 \le r < d.,$ For remainders $,r,r',$ in that interval $\ r\equiv r'\iff r = r',$ since their difference has magnitude $< d$ so is divisible by $,d,$ iff the difference $= 0.,$ Alternatively one uses the general criterion $,a\equiv b\iff d\mid a-b.\ $ – Bill Dubuque Apr 04 '15 at 20:33
-
if you have numbers such as $7+2+4+5$ can you add them all up and then find the remainder because they are small integers and easy to calculate? – Csci319 Apr 04 '15 at 20:35
-
@Csci319 Yes, you can delay reducing them, and reduce them in any order convenient. As long as you always replace the arguments by congruent arguments, the expression remains congruent to the original.so its remainder stays the same, since $,a\equiv b\pmod d \iff (a\ {\rm mod}\ d) = (b\ {\rm mod}\ d)\ $ Generally congruences are more flexible than mod operators. So do everything with congruences, and, only at the end, take remainders (if need be). – Bill Dubuque Apr 04 '15 at 20:38
1
HINT: $111=10\cdot11+1\equiv1\pmod{11}$, and $10\equiv-1\pmod{11}$, so $-10\equiv1\pmod{11}$.
Brian M. Scott
- 616,228