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I was study Babel Civilization ( Babylon ) , I see what about the number $60$ now I'm going to find a trick or fast method to know the rest of divisibility by number $60$ for example :

$6689=60^{2}+51.60+29$

$2567=42.60+47$

I know that the number $60$ divisible by $1,2,3,4,5,6$

So I need fast know method I mean how the rest ? I don't if we can generalized or no ? Please I need fast method without using calculator !

For example $17894=?$

See that : $2177,12=2177+0,12$

$2177=36.60+17$ $0,12=\frac{3}{25}=\frac{60.3}{25.60}=?$ I need written in base $60$ ?

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    $\begingroup$ What do you want to know: when a number is divisible by $60$ or how to find a representation in base $60$? $\endgroup$
    – Qi Zhu
    Commented Dec 30, 2019 at 9:03
  • $\begingroup$ Yes the base and without calculator ? $\endgroup$ Commented Dec 30, 2019 at 9:07

4 Answers 4

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Being divisible to $60$ is equivalent to being divisible simultaneously to $3,20$. So we can use divisibility test for these, which are easy in base $10$. Divisibility by $20$ mean the last $2$ digits must be and even number followed by $0$. Divisibility by $3$ mean sum of digits is divisible by $3$.

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    $\begingroup$ Read closely: OP wants the remainder ("rest"), not simply a divisibility test, and apparently also a way to compute the radix $60$ rep. $\endgroup$ Commented Dec 30, 2019 at 21:25
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Here a method to find the rest of division of $N$ by $60$

1)Finding the rest for $10^n$ we put $t=40$ and $x=60$

$$10^2=x+t\\10^3=16x+t\\10^4=2x^2+46x+t\\10^5=27x^2+46x+t\\10^6=4x^3+37x^2+46x+t\\......\\......\\10^n=Mx+t\space \text{for }n\ge3 $$ It follows a method we apply here particularly to the number $17894$ proposed by the O.P. $$17894=Nx+(1+7+8)t+94=Nx+16\cdot40+94=Nx+734$$ it is easy to verify now that $17894=Nx+734\equiv734\equiv14\pmod{60}$

Method.-If $N=a_na_{n-1}\cdots a_2a_1a_0\equiv x\pmod{60}$ then $x\equiv 40(a_n+\cdots a_2)+a_1a_0$ so if $a_n+\cdots +a_2\equiv h\pmod3$ then $$\boxed{x\equiv 40h+a_1a_0\pmod{60}}$$ where $h$ is equal to $0,1$ or $2$.

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If I understand correctly then you want a quick way to convert a number into base $60$, without using a calculator.

The standard way of converting a number into binary by a computer is to subtract $1$ if it is odd, then divide by $2$ repeatedly, until you get to $0$. We can generalise this method to base $60$.

With your example, $6689$, you can follow this method. You have to find the largest number less than this which is divisible by $60$. In this case, it is $6660$. So $6689=6660+29$. Then you can divide the number by $60$, $6689=60(111)+29$. Now we can do the same for $111$. We want the largest number less than $111$ which is divisible by $60$. In this case, $111=1(60)+51$. So we can write, $6689=60(60(1)+51)+29$. Now we have got down to $1$ in the middle brackets so we can stop. And so $6689$ in base $60$ would be $1,51,29$.

The only question now is how to find the largest number less than your number which is divisible by $60$. You can either do this by guessing and intuition, or use modular arithmetic (if you know about modular arithmetic then read the next paragraph).

For example, we can use $6689$ again. We want the number modulo $2,3,10$ (the remainder when dividing by these numbers). $6689$ is odd so it is congruent to $1$ mod $2$. $6689$ has a digit sum of $29$, which is $2$ more than a multiple of $3$, so $6689$ is congruent to $2$ mod $3$. Last digit of $6689$ is $9$, so $6689$ is congruent to $9$ mod $10$. We can use this information to find $6689$ mod $60$. You can use Chinese remainder theorem, or just use some logic. First we find it mod $6$. We want the number from $0$ to $6$, which is odd, and $2$ more than a multiple of $3$, which is $5$. So $6687$ is congruent to $5$ mod $6$. Now we want a number from $0$ to $60$ which is $5$ more than a multiple of $6$ and ends in a $9$. This is $29$. So $6689$ is congruent to $29$ mod $60$. So we subtract $29$ to get $6660$ is divisible by $60$.

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For a number to be divisible by a composite number, it should be divisible by its individual prime factors raised to their highest powers. like prime factorization of 60 is [2,2,3,5].

so, 60 = (2^2)*3*5 = 4*3*5.

Now, we have to make sure that the number is divisible by $3, 4, 5$. For, a number to be divisible by $5$, the last digit should be either $0$ or $5$. For, a number to be divisible by $4$, the last two digits should be divisible by $4$. Hence, here for a number to be divisible by $4$ & $5$, the last digits should be $0$, and the second last digits should be even.

Next, for a number to be divisible by $3$, the sum of digits should be divisible by $3$.

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