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.

Is this a homework problem? I assume that you mean "begin with a 6" in base 10 representation. — Geoff Robinson, May 23 '15 at 14:18
@GeoffRobinson: Of course it's base $10$! I notice that you didn't query whether $25$ was in base $10$. That's because base $10$ is understood, unless otherwise specified. — TonyK, May 23 '15 at 14:29
@TonyK To be an absolute pedant about this, this question could be in almost any base whatsoever that has 2, 5, 6 among the digits. Base 10 is a very reasonable guess, but in the absence of Tom confirming, it's still just a guess. — Robert Soupe, May 23 '15 at 18:32
@RobertSoupe: Base $10$ is not just a "very reasonable guess", it's the default assumption. (And if it's pedantry that you want, the post has to be in base $10$, doesn't it? The only question is what does $10$ mean.) — TonyK, May 23 '15 at 19:22
@TonyK I somehow expected that last part to come from Geoff rather than from you. — Robert Soupe, May 24 '15 at 04:27

score 7 · Accepted Answer · answered May 23 '15 at 14:19

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.

score 2 · Answer 2 · answered May 23 '15 at 14:18

2

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

answered May 23 '15 at 14:18

Crostul

36,738
4
36
72

score 0 · Answer 3 · answered May 23 '15 at 18:35

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).

score 0 · Answer 4 · answered May 23 '15 at 19:14

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...

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

4 Answers4