Skip to main contentSkip to solution

If 57≡x( mod 5)57 \equiv x (\bmod 5), then the least positive value of xx is:

Solution

✅ Correct Option: 4

57=11×5+257 = 11 \times 5 + 2. So 57≡2(mod5)57 \equiv 2 \pmod{5}. Least positive value of x=2x = 2.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question