The remainder when 2832^{83}283 is divided by 9 is 5.

Quick Scoop: Core Idea

To find what is the remainder when 2^83 is divided by 9 , notice that powers of 2 repeat in a cycle when taken modulo 9:

  • 21=2⇒2mod 9=22^1=2\Rightarrow 2\mod 9=221=2⇒2mod9=2
  • 22=4⇒4mod 9=42^2=4\Rightarrow 4\mod 9=422=4⇒4mod9=4
  • 23=8⇒8mod 9=82^3=8\Rightarrow 8\mod 9=823=8⇒8mod9=8
  • 24=16⇒16mod 9=72^4=16\Rightarrow 16\mod 9=724=16⇒16mod9=7
  • 25=32⇒32mod 9=52^5=32\Rightarrow 32\mod 9=525=32⇒32mod9=5
  • 26=64⇒64mod 9=12^6=64\Rightarrow 64\mod 9=126=64⇒64mod9=1

So the pattern of remainders for 2nmod 92^n\mod 92nmod9 is:

2, 4, 8, 7, 5, 1 and then it repeats every 6 powers.

This means the cycle length is 6.

Mini Step-by-Step

  1. We only care about the exponent modulo 6 (because the cycle length is 6).
  2. Compute 83mod 683\mod 683mod6:
    • 6×13=786\times 13=786×13=78
    • 83−78=583-78=583−78=5
    • So 83≡5mod 683\equiv 5\mod 683≡5mod6.
  3. That means:

283mod 9=25mod 92^{83}\mod 9=2^5\mod 9283mod9=25mod9

  1. From the cycle above, 25=322^5=3225=32, and

32÷9=3 remainder 532\div 9=3\text{ remainder }532÷9=3 remainder 5

so 32mod 9=532\mod 9=532mod9=5.

👉 Therefore, the remainder when 2832^{83}283 is divided by 9 is 5.

Quick Forum-Style Summary

When people on math and exam forums tackle “what is the remainder when 2^83 is divided by 9,” they almost always use the repeating cycle of powers of 2 modulo 9 and reduce the exponent 83 modulo 6 to land on the same answer: 5.

Final Answer: 5
Meta description (SEO-style): Learn how to find the remainder when 2^83 is divided by 9 using modular arithmetic and repeating cycles of powers, a common trick in competitive exam and forum discussions.