# Finding how many times one number fits into the other

Is there a way to tell - for example - how many $60$s are there in a number that cannot be divided by $60$, like $183$? I know that we should remove all numbers after the decimal point but how can we tell this mathematically?

• By inspection or some kind of algorithm? – Meow May 16 '13 at 17:13
• Write $183=60\times k + r$ with $0< r < 60$. See here: en.wikipedia.org/wiki/Division_algorithm – Sigur May 16 '13 at 17:13
• What do you mean "remove all numbers after the decimal point"? Your question is confusing. – 6005 May 16 '13 at 17:24
• @Goos, he/she is saying that $183/60=3.05$ and if you take only the integer part you'll have the answer, that is, $3$. – Sigur May 16 '13 at 17:29
• @Goos - I suspect the OP means that one divides the number by 60 on a calculator and takes the integer part of the quotient (at least that's how I interpreted it). I prefer the division algorithm approach as Sigur suggests, but I guess it all boils down to the same thing. – Chris Leary May 16 '13 at 17:32

Trick: Probably the easiest way to do this by hand is do long division on $183 / 60$ but throw away the remainder (or remove all the numbers after the decimal point, if you prefer).
Formula: This is generally written $\left\lfloor \frac{183}{60} \right\rfloor$. Here $\lfloor x \rfloor$ is called the "floor" of $x$ and means the greatest integer less than or equal to $x$.
Definition: Let $a$ and $b$ be integers. The integer quotient of $a$ and $b$ is defined to be the largest integer $k$ such that $a > bk$. (Note that this is well-defined if and only if $b > 0$). In the case of 183 and 60, we want the largest integer $k$ such that $183 > 60k$.