Find all numbers divisible by 25, that begin with 6.

Question

please, help me solve this problem:

Find all numbers divisible by 25, that begin with 6.

Regards.

Answer 1

An integer is divisible by $25$ if and only if the last two digits are $00$, $25$, $50$, or $75$. Therefore, there are four possibilities.

$$\pm6X00,\ \pm6X25,\ \pm6X50,\ \pm6X75$$

where $X$ is any sequence of digits, perhaps empty.

(Don't forget the negative integers as well! Though you could quibble about whether a number like $-625$ starts with a minus sign or with a $6$.)

@GeoffRobinson makes the excellent point that this works for decimal, base 10, representation. This would need to be changed for other number systems.

Answer 2

They are all numbers of the form $$6X\dots XYY$$ where $X\dots X$ could be anything, and $YY=00, 25, 50, 75$.

Answer 3

I'm going to assume base 10 (decimal), and that they have to be positive.

There are infinitely many such numbers. For any $n > 1$, you can just do $6 \times 10^n$ to get one such number, then keep adding 25 and stop short of $7 \times 10^n$ to get all such numbers with $n + 1$ digits.

Do note that $625 = 25^2 = 5^4$. Are there other powers of 5 that begin with 6? That's a slightly more interesting question. (There's at least one other).

Answer 4

If you want use the rules of modular-arithmetic you can write: $$z=a_0+a_1\cdot 10+...+a_n\cdot 10^n$$ With $a_n=6$ The number $z$ is divisible for $25$ if $$a_0+10\cdot a_1$$ is a multiple of $25$ Indeed $$a_0\equiv a_0\pmod {25}$$ $$10\cdot a_0\equiv 10\cdot a_0\pmod {25}$$ While the other digits are divisible for $25$ Indeed $$10^2\equiv 0\pmod {25}$$ And etc...