Number TheoryMediumModular Arithmetic ShareModular Arithmetic · Medium problemWhich of these congruences has NO solution?A14x≡7(mod21)14x \equiv 7 \pmod{21}14x≡7(mod21)B6x≡9(mod15)6x \equiv 9 \pmod{15}6x≡9(mod15)C10x≡4(mod25)10x \equiv 4 \pmod{25}10x≡4(mod25)D8x≡12(mod20)8x \equiv 12 \pmod{20}8x≡12(mod20)Check Show answer Walk through it step by step Practice more like this Review this concept