Integer chosen uniformly from 1 to 100. What is the probability that it is relatively prime to 100?

Introduction / Context:“Relatively prime to 100” means gcd(n, 100) = 1. Euler’s totient φ(100) counts such integers in {1, …, 100}.

Given Data / Assumptions:

• 100 = 2^2 × 5^2.
• φ(n) = n × ∏(1 − 1/p) over distinct prime factors p of n.

Concept / Approach:Compute φ(100) and divide by 100 to obtain the desired probability.

Step-by-Step Solution:φ(100) = 100 × (1 − 1/2) × (1 − 1/5) = 100 × (1/2) × (4/5) = 40.Probability = φ(100)/100 = 40/100 = 2/5.

Verification / Alternative check:Numbers divisible by 2 or 5 are not coprime to 100; inclusion–exclusion gives the same count.

Why Other Options Are Wrong:61/100 overcounts; 17/100 and 19/100 undercount; 3/5 is the complement of 2/5 among multiples of 2 or 5, not the coprime share.

Common Pitfalls:Confusing “not divisible by 2 or 5” with “coprime to 100” (these notions coincide here but must be handled carefully with inclusion–exclusion).

Final Answer:2/5