# How do I know if the number is divisible by $60$?

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 $$60$$ ?

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

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 $$3$$.

• Read closely: OP wants the remainder ("rest"), not simply a divisibility test, and apparently also a way to compute the radix $60$ rep. Dec 30, 2019 at 21:25

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 $$2$$.

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 $$60$$.

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 $$3$$.