Mathematics

Mathematics

Problem

A student claims that the reciprocal of a natural number nn has a terminating decimal expansion of exactly four digits after the decimal point. Investigate this claim as follows: a. (i) Show that this situation is possible only when nn can be expressed in the form 2a5b2^{a}5^{b}, where aa and bb are non-negative integers. b. (ii) Determine all such natural numbers nn for which 1n\dfrac{1}{n} terminates after exactly four decimal places, and state the smallest value of nn satisfying this condition.
Intuitive explanation

The key insight is that a fraction in lowest terms has a terminating decimal expansion if and only if its denominator has no prime factors other than 2 and 5. For 1n\frac{1}{n} to terminate in exactly four decimal places, nn must be expressible as 2a⋅5b2^a \cdot 5^b where the larger exponent determines the number of digits—specifically, max⁡(a,b)=4\max(a,b) = 4. Your job is to find all such natural numbers nn and verify whether the claim is actually possible.

Intuitive explanation

The key insight is that a fraction in lowest terms has a terminating decimal expansion if and only if its denominator has no prime factors other than 2 and 5. For 1n\frac{1}{n} to terminate in exactly four decimal places, nn must be expressible as 2a⋅5b2^a \cdot 5^b where the larger exponent determines the number of digits—specifically, max⁡(a,b)=4\max(a,b) = 4. Your job is to find all such natural numbers nn and verify whether the claim is actually possible.

Browse the problem library