When working modulo m, the notation a^-1 is used to denote the residue b for which ab≡1(modm), if any exists. For how many integers a satisfying 0 < a < 100 is it true that a(a−1)−1≡4a−1(mod20)?
Please hurry up guys!!
+26396 When working modulo m, the notation a−1 is used to denote the residue b for which ab≡1(modm),
if any exists.
For how many integers a satisfying 0<a<100 is it true that a(a−1)−1≡4a−1(mod20) ?
a(a−1)−1≡4a−1(mod20)|⋅aa2(a−1)−1≡4a−1⋅a(mod20)|a−1⋅a=aa=1a2(a−1)−1≡4(mod20)|a2(a−1)−1=a2a−1=a+1+1a−1a+1+1a−1⏟is integer, only if a=2≡4(mod20)|a=22+1+12−1≡4(mod20)|3+1≡4(mod20)4≡4(mod20) ✓
One Integer solution a=2
+26396 When working modulo m, the notation a−1 is used to denote the residue b for which ab≡1(modm),
if any exists.
For how many integers a satisfying 0<a<100 is it true that a(a−1)−1≡4a−1(mod20) ?
a(a−1)−1≡4a−1(mod20)|⋅aa2(a−1)−1≡4a−1⋅a(mod20)|a−1⋅a=aa=1a2(a−1)−1≡4(mod20)|a2(a−1)−1=a2a−1=a+1+1a−1a+1+1a−1⏟is integer, only if a=2≡4(mod20)|a=22+1+12−1≡4(mod20)|3+1≡4(mod20)4≡4(mod20) ✓
One Integer solution a=2
Er... I don't think that it is...
For one, it says integers
Second, my friend has seen this before and he says it's not 1
+2234 Solving it this way gives:
a(a−1)−1≡4a−1(mod20)a≡4a−1(a−1)(mod20)a2≡4(a−1)(mod20)a2−4a+4≡0(mod20)(a−2)2≡0(mod20) a=2+10n| for n = (0, 1, 2, … 9), so 10 integers satisfy 0 < a < 100
Well blimey! You’re right. (I’m surprised because Heureka can do mod problems while napping.)
GA
