Skip to content
Exercise 1 · Q2

Q.Find 76(mod3)7^6 \pmod 3.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
43% · 46/106 Questions
✓ Free question

Since 7≡1(mod3)7\equiv 1\pmod 3, any power of 77 is ≡1\equiv 1, so 76≡1(mod3)7^6\equiv 1\pmod 3.

If a≡r(modm), then an≡rn(modm)\text{If } a \equiv r \pmod m,\ \text{then } a^n \equiv r^n \pmod m

Here a=7, m=3, n=6a=7,\ m=3,\ n=6.

  1. Reduce the base. 7=2×3+1⇒7≡1(mod3)7 = 2\times 3 + 1 \Rightarrow 7 \equiv 1 \pmod 3.
  2. Raise to the power. 76≡16(mod3)7^6 \equiv 1^6 \pmod 3.
  3. Simplify. 16=11^6 = 1, so 76≡1(mod3)7^6 \equiv 1 \pmod 3.
✓Final answer

76≡1(mod3)7^6 \equiv 1 \pmod 3.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.