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
$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$
proof-writing modular-arithmetic congruence-relations
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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
$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$
proof-writing modular-arithmetic congruence-relations
The titled question differs from that in the body. Please correct.
– Bill Dubuque
Nov 14 at 16:14
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up vote
0
down vote
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up vote
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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
$$frac{a}{d} x equiv frac{c}{d} mod frac{m}{d} .$$
proof-writing modular-arithmetic congruence-relations
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
proof-writing modular-arithmetic congruence-relations
edited Nov 14 at 16:04
Davide Giraudo
123k16149253
123k16149253
asked Nov 14 at 15:58
N.Luscomb
1
1
The titled question differs from that in the body. Please correct.
– Bill Dubuque
Nov 14 at 16:14
add a comment |
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
add a comment |
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The titled question differs from that in the body. Please correct.
– Bill Dubuque
Nov 14 at 16:14