If d = gcd(a, m) and d|c, then show that the congruence ax ≡ c (mod m) is equivalent to a x≡c (modm).











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If $d = gcd(a, m)$ and $d|c$, then show that the congruence $ax equiv c$ (mod $m$) is equivalent to



$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$










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  • The titled question differs from that in the body. Please correct.
    – Bill Dubuque
    Nov 14 at 16:14















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down vote

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If $d = gcd(a, m)$ and $d|c$, then show that the congruence $ax equiv c$ (mod $m$) is equivalent to



$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$










share|cite|improve this question
























  • The titled question differs from that in the body. Please correct.
    – Bill Dubuque
    Nov 14 at 16:14













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0
down vote

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up vote
0
down vote

favorite











If $d = gcd(a, m)$ and $d|c$, then show that the congruence $ax equiv c$ (mod $m$) is equivalent to



$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$










share|cite|improve this question















If $d = gcd(a, m)$ and $d|c$, then show that the congruence $ax equiv c$ (mod $m$) is equivalent to



$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$







proof-writing modular-arithmetic congruence-relations






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edited Nov 14 at 16:04









Davide Giraudo

123k16149253




123k16149253










asked Nov 14 at 15:58









N.Luscomb

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1












  • The titled question differs from that in the body. Please correct.
    – Bill Dubuque
    Nov 14 at 16:14


















  • The titled question differs from that in the body. Please correct.
    – Bill Dubuque
    Nov 14 at 16:14
















The titled question differs from that in the body. Please correct.
– Bill Dubuque
Nov 14 at 16:14




The titled question differs from that in the body. Please correct.
– Bill Dubuque
Nov 14 at 16:14















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