WAEC 2019 · Paper 1 · Q5

If 7+y≡4(mod8)7 + y \equiv 4 \pmod 8, find the least value of yy with 10≤y≤3010 \le y \le 30.

Worked solution (try it first)
  1. Take 7 from both sides: y≡4−7=−3(mod8)y \equiv 4 - 7 = -3 \pmod 8.
  2. Add 8 to make it positive: y≡5(mod8)y \equiv 5 \pmod 8, so yy is one of 5, 13, 21, 29, ….
  3. The least of these with 10≤y≤3010 \le y \le 30 is 13, option B.

Report a problem with this question