Find the remainder when $ 528528528...$up to $528$ digits is divided by $27$?

Here is a python3 session

>>> s = '528' * 176
>>> len(s)
528
>>> int(s) % 27
21

You can see that $6$ cannot be correct by casting out $9$'s: Since $5+2+8=5+5+5$, we have

$$528528\ldots528\equiv5+5+5+\cdots+5+5+5=5\cdot528\equiv5(5+2+8)\equiv5\cdot6\equiv3\mod 9$$

so the remainder mod $27$ must be either $3$, $12$, or $21$. Your approach gave the correct answer, $21$.

Tags:

Elementary Number Theory

Divisibility

Related

Solve: $\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}} =4 $ Show that there exist $n$ such that $r$ divides $\binom{p^n}{q^n}$ will a nonempty countable and compact subset of a metric space always contain an isolated point? (1) Sum of two factorials in two ways; (2) Value of $a^{2010}+a^{2010}+1$ given $a^4+a^3+a^2+a+1=0$. Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Does there exist an injection $f:\mathbb{R}\to P(\mathbb{N})$ that satisfies these conditions? Cantor Set (Hausdorff) $f:\mathbb{R}\rightarrow\mathbb{R}$ is not one-to-one if $[f(x)]^2-4\cdot f(x^5)+3=0,\forall x\in\mathbb{R}$? How to use Stokes theorem in the presence of holes? Let $T$ be any subset of $\{1,2,3,...,100\}$ with $69$ elements. Prove that one can find four distinct integers such that $a+b+c=d$. Prove that $\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}$ Is there a *simple* example showing that uncorrelated random variables need not be independent?

Recent Posts