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How to find the value of '?'
a mod m = b mod m , will this formula be used?
I am taking discrete maths course for CS. And this question is from one of its chapter

rdr2
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    What is $438 (\mod 11)$? And what is the order of $(\Bbb Z / 11 \Bbb Z)^*$? – Robert Shore Mar 21 '19 at 06:39
  • 438 mod 11 would be 9 And I don't know what you mean by order of (Z/11Z) – rdr2 Mar 21 '19 at 06:43
  • For us to be able to help you effectively, you need to provide more background. Why is this problem of interest to you? How have you tried to solve it? – Robert Shore Mar 21 '19 at 06:44
  • I am taking discrete maths course for CS. And this question is from one of its chapter – rdr2 Mar 21 '19 at 06:47
  • Thanks. How have you tried to solve it? You should edit this information directly into your question, rather than using a comment. – Robert Shore Mar 21 '19 at 06:49
  • Have you taken any classes that cover modulo arithmetic. Fermat's Little Theorem basically is the first thing you should thing of. ANd it should be the last and it gives you the answer immediately with no surprises. – fleablood Mar 21 '19 at 06:59
  • There are two relationships, a mod m = b mod m. And gcd{a,m} = gcd {b,m} Maybe either of these two can be used to find the value of b, but I don't understand how to apply them or how to solve for the power – rdr2 Mar 21 '19 at 07:01
  • @fleablood I have taken modulo arithmetic classes, but we haven't covered fermat's little theorem yet – rdr2 Mar 21 '19 at 07:01
  • Well what have you tried. Do you know what $438\pmod {11}\equiv k \pmod{11}$ is? Have you tried seeing what $k^2, k^3, k^4 \pmod {11}$ are and seeing if there is any pattern? – fleablood Mar 21 '19 at 07:04
  • would k be 20 ? if we solve for 438 (mod 11) = k (mod 11)? – rdr2 Mar 21 '19 at 07:12
  • also i looked at fermet's little theorem. For that should the power be a prime number? but in this case the power is divisible by 7 – rdr2 Mar 21 '19 at 07:26
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    Possible duplicate of How do I compute $a^b,\bmod c$ by hand? – Arnaud D. Mar 21 '19 at 10:47

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Observe that $\text {gcd} (11,438) = 1.$ So by Euler's theorem $438^{10} \equiv 1\ (\text {mod}\ 11).$ So $438^{87490} \equiv 1\ (\text {mod}\ 11).$ Also $438 \equiv -2\ (\text {mod}\ 11) \implies 438^3 \equiv -8 \equiv 3\ (\text {mod}\ 11).$ Therefore $$438^{87493} \equiv 3\ (\text {mod}\ 11).$$

If you don't know Euler's theorem or Fermat's little theorem then observe from my above calculation that $438^3 \equiv 3\ (\text {mod}\ 11).$ So $438^6 \equiv 9 \equiv -2\ (\text {mod}\ 11) \implies 438^{30} \equiv (-2)^5 \equiv -32 \equiv 1 (\text {mod}\ 11).$

Observe that $87493 = 87480 + 13.$ Since $30 \mid 87480$ so $438^{87480} \equiv 1\ (\text {mod}\ 11).$ Since $438^6 \equiv -2\ (\text {mod}\ 11)$ so $438^{12} \equiv 4\ (\text {mod}\ 11).$ Again $438 \equiv -2\ (\text {mod}\ 11).$ So $438^{13} \equiv 438^{12} \cdot 438 \equiv 4 \cdot (-2) \equiv -8 \equiv 3\ (\text {mod}\ 11).$ Thus we get

$(1)$ $438^{87480} \equiv 1\ (\text {mod}\ 11).$

$(2)$ $438^{13} \equiv 3\ (\text {mod}\ 11).$

Therefore what is $438^{87493} \equiv ~?\ (\text {mod}\ 11)$?

$$438^{87493} \equiv 438^{87480+13} \equiv 438^{87490} \cdot 438^{13} \equiv 1 \cdot 3 \equiv 3\ (\text {mod}\ 11).$$

little o
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