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What is the easiest way to know if large number is divisible by 57? For example, how could I deduce that 57 divides 300000177?

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Do you mean if its divisible by 57? Its remainder? The present statement is unclear; the question should be edited to clarify this. (You'll also want to switch the tag from number-theory to elementary-number-theory; the former isn't really suitable. The divisibility and modular-arithmetic tags may also be advisable.) – Semiclassical Aug 19 '14 at 14:02
@Semiclassical : the same ideas but for 57 – Kumar Aug 19 '14 at 14:04
Seems it must be "divisible by 57", and that the use of "divided" in the question can be explained by language problem. – coffeemath Aug 19 '14 at 14:05
thanks @coffeemath – Kumar Aug 19 '14 at 14:06
Note that $57=3\cdot 19$. It's easy to tell if a number is divisible by $3$ (sum its digits, sum the digits of that, etc. until you can tell whether you're getting multiples of three). So the only tricky thing is finding a divisibility rule for 19. Also, note that the Chinese remainder theorem is applicable for dealing with the remainder of division by 57. – Semiclassical Aug 19 '14 at 14:10
up vote 5 down vote accepted

As noted in the comments, $n$ is divisible by $57$ if and only if $n$ is divisible by $3$ and by $19$.

Divisibility test for $3$ is well known (as noted by Semiclassical above): Just see if the sum of the digits is divisible by $3$.

Divisibility test for $19$: Take the last digit of $n$ and double it. Add this result to what is left of $n$ after removing the last digit. Your original number is divisible by $19$ if and only if your final answer is divisible by $19$.

Reasoning: Write $n=10a+b$ where $b$ is a $1$-digit number. We have: $$19\mid 10a+b \Leftrightarrow 19\mid 10a+b+19b \Leftrightarrow 19\mid 10a+20b\Leftrightarrow 19\mid a+2b$$ (last follows since $10$ and $19$ are relatively prime).

For longer numbers, apply the rule as many times as necessary.

Example: $n=2137$. $213+2\cdot 7=227$. $22+2\cdot 7=36$. So $2137$ is not divisible by $19$.

(Note: Divisibility tests here assume that $n$ is expressed in base 10).

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To speed up a little, you can chop off the last two digits, and multiply by 4 instead. – Michael Aug 19 '14 at 15:25

To check divisibility by $57$ you simply check divisibility by $3$ and $19$.

If you are familiar with modular arithmetic, use the fact that

$$20 \equiv 1 \pmod{19}$$ Therefore, if you have a number $n$ add the last 2 digits with 5 times the rest and repeat. $n$ is divisible by $19$ if and only if any of the numbers you get is divisible by $19$.

$$300000177 \rightarrow 77+5*3000001=15000082 \to 82+5*150000=750082 \to 82+5*7500=37582 \to 82+5*375=1957 \to 57+5*19$$ as both 57 and 19 are divisible by 19, the original number is also divisible by 19.

If you want directly a divisibility rule by $57$, use that $399$ is a multiple of 57. Therefore $$400 \equiv 1 \pmod{57}$$

this leads to a similar more complicated rule.

Added $20 \equiv 1 \pmod{19}$ means $100 \equiv 5 \pmod{19}$. Write $n =100 a+b$. Then $$n =100a+b \equiv 5a+b \pmod{19}$$

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This is nice, but can you elaborate a little more how the fact that $20 \equiv 1 \pmod {19}$ leads to that divisibility criterion? – Ant Aug 19 '14 at 14:37
@Ant Done...... – N. S. Aug 19 '14 at 14:58
Easier: $\ 7)\!\underset{\equiv 14}2\!\!+7)\underset{\equiv 4}2\!+1)\!\underset{\equiv 10}2\!\!+0)\underset{\equiv 1}2\!+0)\underset{\equiv 2}2\!+0)\underset{\equiv 4}2\!+0)\underset{\equiv 8}2\!+0)\!\underset{\equiv 16}2\!\!+3\equiv 0\pmod{19},\,$ which uses only very simple mental arithmetic (doubling and adding a single digit, $ $ mod $\,19),\,$ see my answer. – Bill Dubuque Aug 19 '14 at 15:32

$5+7=12=3\cdot4$, so divisibility by $57$ implies being a multiple of $3$. Since $3+1+7+7=18=$ $=3\cdot6$, your number meets this demand, the other being divisibility by $19$. Since $100=5\cdot20=$ $=5(19+1)$, your number is divisible by $19$ if $77\cdot5^0+01\cdot5^1+00\cdot5^2+00\cdot5^3+3\cdot5^4=77+$ $+5+3\cdot625=82+1875=1957=19\cdot103$ is a multiple of $19$, which is indeed the case.

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Hint $\ \ 3\mid n\iff\,$ the sum of the decimal digits of $\,n\,$ is $\,\equiv 0\pmod 3,\,$ by $\,10\equiv 1\pmod 3$

and $\ \ \ 19\mid n\iff 19\,$ divides the reversed digit number in radix $2\,$ (binary), $ $ since, e.g.

$\!\!\!\begin{eqnarray} &&19\mid\ \ d_3 10^3\! + d_2 10^2\! + d_1 10 + d_0\\ \iff &&19\mid (d_3 10^3\! + d_2 10^2\! + d_1 10 + d_0)\, 2^3\ \ {\rm by}\ \ (19,2) = 1. \ \text{ Thus, using} \ \, 20\equiv 1\!\!\!\pmod{19}\\ \iff &&19\mid\ d_3 + d_2 2 + d_1 2^2 + d_0 2^3 = \text{ reversed number in binary}\\ \iff &&19\mid\ d_3 + 2(d_2 + 2(d_1 + 2 d_0))\ \ \text{in Horner form} \end{eqnarray}$

e.g. $\quad\! 19\mid 5016\ $ by $\ 6)\!\underset{\equiv\color{#0a0}{ 12}}{2}\!\!\color{#c0F}{+1})\underset{\equiv \color{#c00}7}2\!+ 0)\!\underset{\equiv 14}2\!+5\equiv 0\pmod{19}\ $ takes only $\,5\,$ secs mentally.

$\equiv n$ is partial eval: $\,(6)2\equiv \color{#0a0}{12},\,$ next $\,(\color{#0a0}{12}\!\color{#c0f}{+\!1})2\equiv\color{#c00} 7,\ \ldots$

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Number is divisible by 57 if the sum of '4 times hundreds + last two digit' is divisible by 57. For more detail about cross divisibility test refer Divisibility criteria for $7,11,13,17,19$

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