This is for my assembly language class. I am finding different answers. My answer was $15_7$. But a friend got 2824. Can someone please explain the correct way to do it if $15_7$ is wrong?
$$(235432_7 \cdot 2551_7) \pmod{311_7} = N_7 = ?$$
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Wolfram alpha (online) will tell you the correct answer. – Ethan Bolker Dec 22 '15 at 18:05
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This site uses MathJax formatting for formulae. Also, questions such as this could easily be answered by calculators such as Wolfram-Alpha without the need to post. – JMoravitz Dec 22 '15 at 18:05
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For clarification, are you asking for $2354327\cdot 25517\pmod{3117}$ where these numbers are in decimal or are you asking for $235432\cdot 2551\pmod{311}$ where these numbers are in base 7? – JMoravitz Dec 22 '15 at 18:09
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It is in base 7 – groot Dec 22 '15 at 18:09
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not decimal, so the calculator wouldn't work i don't think – groot Dec 22 '15 at 18:10
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Well, your friend was correct in base 10. But if this is supposed to be in base 7... (um, how were we supposed to know that?) – fleablood Dec 22 '15 at 18:14
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there is a 7 subscript... – groot Dec 22 '15 at 18:17
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Then you should fix the title. Your friend did the question in the title which is a completely different question altogether. – fleablood Dec 22 '15 at 18:18
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I got 15 base 7 using the calculator and converting bases to decimal and back to base 7 – groot Dec 22 '15 at 18:18
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Well, I did it by hand and got error after error. We can conclude your friend misunderstood the question and did it with trailing 7s in base 10. You seem to be doing it right. Unlike me. – fleablood Dec 22 '15 at 18:48
2 Answers
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$235432$ in base 7 is $2+3\cdot 7 + 4\cdot 7^2+5\cdot 7^3+3\cdot 7^4+2\cdot 7^5 = 42751$ in base 10
$2551$ in base 7 is $1+5\cdot 7+5\cdot 7^2+2\cdot 7^3 = 967$ in base 10
$311$ in base 7 is $1+1\cdot 7 + 3\cdot 7^2 = 155$ in base 10
$42751\cdot 967\pmod{155}\equiv 12$
$12$ in decimal is $5+1\cdot 7$ is $15$ in base 7
Note, if the answer was expected to be in base 7, then your friends answer of $2824$ doesn't even make sense since $8$ is not a valid digit in base 7.
JMoravitz
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I'm going to use notation $235432$ as $2.3.5.4.3.2$ where it's understood this is a sum and each term between periods is multiplied by a power of seven.
$2.3.5.4.3.2 \equiv 17.5.4.3.2 - 15.5.5.0.0\equiv 2.0.-1.3.2 \equiv 1.6.6.3.2 \equiv 13.6.3.2 - 12.4.4.0 \equiv 1.2.-1.2 \equiv 8.6.2 - 6.2.2 \equiv 2.4.0 \equiv 240_7 \mod (311_7)$
$2551_7 \equiv 2.5.5.1 \equiv 19.5.1 - 18.6.6 \equiv 1.-1.-5 \equiv 6.-5 \equiv 5.2 \equiv 52_7 \mod (311_7)$
$240_7*52_7 \equiv 10.20+4.8.0 \equiv 10.24.8.0 - 9.3.3.0 \equiv 1.21.5.0 \equiv 28.5.0 - 27.9.9 \equiv 1.-4.-9 \equiv 3.-9 \equiv 1.14-9 \equiv 1.15 \equiv 15_7 \mod (311_7)$
fleablood
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