I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 31) \equiv 7 (\mod 31)$, etc.). While this makes the computation easier, I'm thinking there might be a better way to do this.
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How to efficiently compute $17^{23} (\mod 31)$ by hand?
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$17^2\equiv10\pmod{31}$
$17^4 \equiv 7\pmod{31}$
$17^3 \equiv 170\pmod{31}\equiv15\pmod{31}$
$17^7=17^3\times17^4\equiv15\times7\pmod{31}\equiv12\pmod{31}$
${(17^7)}^3=17^{21}\equiv23\pmod{31}$
$17^{23}=17^{21}\times17^2\equiv230\pmod{31}\equiv13\pmod{31}$
answered Oct 30 at 4:18
Shobhit
3,4711632
3,4711632
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Note that $17\equiv 3/2 \pmod {31}$. We can then calculate this.
$2^5 = 1$, so $2^{23}=8$.
$3^3 = -4$, so $3^6 = 16$. We then have $3^{24} = 16^4 = 2$, and dividing through by $8$, we get $8$. Divide this by $3$ to get $31+8 = 39$, divide by 3 to get $13$.
answered yesterday
wendy.krieger
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2,3391311
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How to efficiently compute $17^{23} (\mod 31)$ by hand?
up vote
14
down vote
I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 31) \equiv 7 (\mod 31)$, etc.). While this makes the computation easier, I'm thinking there might be a better way to do this.
asked Oct 30 at 4:02
pldimitrov
696
696
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$17^2\equiv10\pmod{31}$
$17^4 \equiv 7\pmod{31}$
$17^3 \equiv 170\pmod{31}\equiv15\pmod{31}$
$17^7=17^3\times17^4\equiv15\times7\pmod{31}\equiv12\pmod{31}$
${(17^7)}^3=17^{21}\equiv23\pmod{31}$
$17^{23}=17^{21}\times17^2\equiv230\pmod{31}\equiv13\pmod{31}$
answered Oct 30 at 4:18
Shobhit
3,4711632
3,4711632
@FrenzYDT. Thanks for the edit. - Shobhit Oct 30 at 4:34
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