400 and 1s

Call our mystery base $b$ . We need to solve $1b^3 + 1b^2 + 1b^1 + 1b^0 = 400$ , i.e. $b^3 + b^2 + b = 399$ , therefore $b$ divides 399. Since $399 = 3 \times 7 \times 19$ , we are left with only a few possible values of $b$ :

$b=1 \Rightarrow b^3 + b^2 + b + 1 = 4 \Rightarrow$ no solution

$b=3 \Rightarrow b^3 + b^2 + b + 1 = 40 \Rightarrow$ no solution

$b=7 \Rightarrow b^3 + b^2 + b + 1 = 400 \Rightarrow\ b = \boxed{7}$ .

Higher values of $b$ are not possible since they will certainly yield $b^3 + b^2 + b + 1 > 400$ .

Call our mystery base $x .$ $1000 < 1111,$ and $1000$ in base $x$ is $1x^3 + 0x^2 + 0x + 0 = x^3 .$

In base 10, then, $x^3 < 400 .$ This means $x < \sqrt[3]{400} .$ The cube root of 400 is approximately 7.4, so $x$ must be 7 or smaller.

Note also that $1333 > 1111 ,$ and $1331$ in base $x$ is $1x^3 + 3x^2 + 3x + 1 = (x+1)^3 .$ In base 10, $(x+1)^3 > 400,$ so $x + 1$ must be greater than approximately 7.4, implying $x$ is greater than approximately 6.4.

The only choice for $x ,$ then, is 7. Checking 7 itself and converting to base 10, $(1)7^3 + (1)7^2 + (1)7 + (1)1 = 400 .$

Like this solution? Check out more types like this of this at the wiki Diophantine Equations - Solve by Bounding Values .

I did somewhat like Tijemen:

Calling our mystery base b, we're requested to solve $1{b}^{3}+1{b}^{2}+1{b}^{1}+1{b}^{0}=400$ , i.e. ${b}^{3}+{b}^{2}+{b}^{1}= b({b}^{2}+{b}^{1}+1)=399$ , and therefore, b divides 399, which is a composite number: $399={3} * {7} * {19}$ .

Since b is a natural number greater than 1, $3{b}^{3}>{b}^{3}+{b}^{2}+{b}^{1}=399>3{b}^{2}$ , therefore, $5<b<11$ . So, we are left with only one possibility: $b=7$ .

I solved $x^3+x^2+x-399=0$ in a CAS (computer algebra system) I am writing in Ruby with Newton's Method.