What is the least natural number #n# for which the equation #floor(10^n/x)=2018# has an integer solution?

2 Answers
Oct 13, 2017

#7#

Explanation:

Note that #2018# has #4# significant digits, so we would expect #10^n# to be approximately the product of #2018# and another #4# digit number.

Try:

#floor(10^7/2018) = 4955#

Then:

#floor(10^7/4955) = 2018" "# as required

Is there any smaller #n# than #7#?

#floor(10^6/495) = 2020#

#floor(10^5/49) = 2040#

#floor(10^4/4) = 2500#

So #n=7# is the smallest solution.

Oct 13, 2017

See below.

Explanation:

This equation leads to the following relationship

#2017 lt 10^n/x lt 2019#

or equivalently

#10^n/2019 < x < 10^n/2017#

now for
#n = 6->495.295 < 495.786# no solution
#n=7-> 4952.95 < 4953 < 4954 < 4955 < 4956 < 4957 < 4957.86# there are #5# solutions.

so the integer solutions are #x = { 4953,4954, 4955, 4956 ,4957}# for #n = 7#