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# number theory

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Given that $$13^{-1} \equiv 29 \pmod{47}$$, find $$34^{-1} \pmod{47}$$, as a residue modulo 47. (Give a number between 0 and 46, inclusive.)

Jan 1, 2019

#1
+5776
+2

$$13^{-1}\equiv 29 \pmod{47}\\ (29)(13) \equiv 1 \pmod{47}\\ (29)(47-34)\equiv 1 \pmod{47}\\ (29)(47)+(-29)(34)\equiv 1 \pmod{47}\\ -(29)(34) \equiv 1 \pmod{47}$$

$$34^{-1} \equiv (-29) \pmod{47}\\ (-29)\equiv (47-29) \pmod{47}\\ (-29)\equiv 18 \pmod{47}\\ 34^{-1} = 18 \pmod{47}$$

$$\text{as a check}\\ 18\cdot 34 = 612 = 13\cdot 47 + 1\\ 18 \cdot 34 \equiv 1 \pmod{47}$$

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Jan 2, 2019

#1
+5776
+2

$$13^{-1}\equiv 29 \pmod{47}\\ (29)(13) \equiv 1 \pmod{47}\\ (29)(47-34)\equiv 1 \pmod{47}\\ (29)(47)+(-29)(34)\equiv 1 \pmod{47}\\ -(29)(34) \equiv 1 \pmod{47}$$

$$34^{-1} \equiv (-29) \pmod{47}\\ (-29)\equiv (47-29) \pmod{47}\\ (-29)\equiv 18 \pmod{47}\\ 34^{-1} = 18 \pmod{47}$$

$$\text{as a check}\\ 18\cdot 34 = 612 = 13\cdot 47 + 1\\ 18 \cdot 34 \equiv 1 \pmod{47}$$

Rom Jan 2, 2019