5
$\begingroup$

Find the remainder when $528528528...$up to $528$ digits is divided by $27$?
Here's what I have done: The number can be written as $528\cdot 10^{525}+528\cdot 10^{522}+...+528$ which has $176$ terms and each term is $\equiv15 \mod 27$ thus the number should be $176*15 \mod 27$ hence $21$ should be the remainder. But book says it is $6$. I don't understand the flaw in my logic. Please correct me.

|cite|improve this question
$\endgroup$
  • $\begingroup$ you have $21+6=27$ perhaps you are off by a sign? $\endgroup$ – gt6989b Oct 16 '17 at 15:26
  • 3
    $\begingroup$ I think that your answer is correct. $\endgroup$ – alexp9 Oct 16 '17 at 15:29
  • $\begingroup$ Brute force in python gives 21. $\endgroup$ – Gribouillis Oct 16 '17 at 15:31
  • 1
    $\begingroup$ Are you sure the book says $6$ and not $-6$? After all, $21\equiv -6 \pmod{27},$ so then both answers would agree. $\endgroup$ – David K Oct 16 '17 at 15:36
  • 1
    $\begingroup$ @fleablood Also a good point about negative remainders. It looks like there's probably an error in the book. (Unless the transcription of the problem is very confused, it's only $176$ "copies" of the group of digits $528,$ which makes $528$ digits altogether since each group has three digits. Also, we don't need $10^k$ to be congruent to $1$; we only need $10^{3k}\equiv 1 \pmod{27},$ which is true.) $\endgroup$ – David K Oct 16 '17 at 17:00
3
$\begingroup$

Here is a python3 session

>>> s = '528' * 176
>>> len(s)
528
>>> int(s) % 27
21
|cite|improve this answer
$\endgroup$
  • $\begingroup$ Is it possible to multiply a string by a number? My god...python is awesome. $\endgroup$ – Integral Oct 16 '17 at 15:35
  • 2
    $\begingroup$ @Integral yes strings, lists and tuples can be repeated by multiplying them by a number. $\endgroup$ – Gribouillis Oct 16 '17 at 15:36
2
$\begingroup$

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, $21$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Anyway it is also $-6$ so there is a typo in the book or Anuran has not have seen the minus sign. $\endgroup$ – Piquito Oct 16 '17 at 15:54
  • $\begingroup$ @Piquito, I agree, a negative sign would fix things (as David K noted in comments). But remainders are usually understood to be nonnegative, so I'm inclined to think it's a typo. $\endgroup$ – Barry Cipra Oct 16 '17 at 15:57
1
$\begingroup$

Since $3\mid111$, we know that $27\mid999$, Therefore, $$ 1000\equiv1\pmod{27} $$ Thus, $$ \begin{align} \sum_{k=0}^{175}528\cdot1000^k &\equiv528\cdot176\pmod{27}\\ &\equiv3\cdot176^2\pmod{27}\\ &\equiv3\cdot14^2\pmod{27}\\ &\equiv3\cdot7\pmod{27}\\ &\equiv21\pmod{27} \end{align} $$

|cite|improve this answer
$\endgroup$
0
$\begingroup$

You are incorrect in assuming $10^{k} \equiv 1 \mod 27$. As $10^k \not \equiv 1$ we do not have $528*10^k \equiv 15 \mod 27$.

What you need instead is $528528... = 528(1001001001......)$

And $1001001..... =\sum_{k=0}^{175} 10^{3k}$

$10^3 \equiv 1 \mod 27$... Oh... we do have that and you were not wronng after all.... so $\sum 10^{3k}\equiv 176 \equiv 14 \mod 27$.

So $528528....... \equiv 15*14 \equiv 21 \mod 27$.

And... the book is wrong.

Had it been 527 iterations of 528 the answer would be $6$.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

Note that $27 \times 37 = 999$.

To find the remainder you get when you divide $528528\cdots 528$ by $999$, you can "cast out" $999$s.

$\underbrace{528 + 528 + \cdots 528}_{\text{$176$ times} } \to 176 \times 528 \to 92928 \to 92+928 \to 1020 \to 21$

So the remainder is $21$.

Note. If the remainder was bigger than 26, you would have had to divide it by 27 to get the correct remainder.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.