Question

after how many decimal places will the decimal expansion of 23457 upon 2 exponent 3 into 5 exponent 4 terminate​

Answer 1

Answer:

Let n=2α⋅5β⋅m, where gcd(m,10)=1.

The decimal expansion of the rational number an, gcd(a,n)=1 terminates if and only if m=1. The number of digits after the decimal equals γ=max{α,β}.

A rational number an, gcd(a,n)=1, has a terminating decimal (say, with k digits after the decimal) precisely when an⋅10k is and integer but an⋅10k−1 isn’t. Thus, we seek the existence of a positive integer k for which n∣a⋅10k (and the least such equals the number of digits after decimal). In particular, we have m∣a⋅10k. But gcd(m,a)=1 (because gcd(n,a)=1) and gcd(m,10)=1. Therefore, m∣a⋅10k is possible only when m=1 regardless of k.

We have established the decimal expansion of an is terminating if and only if m=1.

We now show that the decimal expansion of a2α⋅5β has γ=max{α,β} digits after the decimal.

Indeed,

a2α⋅5β⋅10γ=a⋅2γ−α⋅5γ−β∈Z,

and

a2α⋅5β⋅10γ−1=a⋅2γ−α−1⋅5γ−β−1∉Z.

The first statement is true because γ−α≥0 and γ−β≥0. The second statement is true because at least one of γ−α, γ−β equals 0 and a is divisible by neither 2 nor 5.