# Thread: Are these base 15 numbers divisible by each of 19 and 25?

1. ## Are these base 15 numbers divisible by each of 19 and 25?

Are these base 15 numbers divisible by each of 19 and 25?

1. CA36D5
2. 1E959B3

I'm using this table here Addition and Multiplication Tables in Various Bases for base 15 and it uses the symbols A, B, C ,D ,E to represent the numbers written in base ten as 10, 11, 12, 13, 14 respectively

One hint from this exercise is to consider divisibility by factors (not necessarily prime) of 19 and 25
I'm allowed to convert the base 15 numbers 19 and 25 to base 10 (19 and 25 are also written in base 15) to find their factors

2. Hello, nacknack!

Are these base-15 numbers divisible by each of $\displaystyle 19_{15}$ and $\displaystyle 25_{15}$ ?

. . $\displaystyle (1)\; CA\,3\,6\,D\,5_{15} \qquad\qquad (2)\; 1E\,9\,5\,9\,B\,3_{15}$

Hint: consider divisibility by factors (not necessarily prime) of $\displaystyle 19_{15}$ and $\displaystyle 25_{15}$

I'm allowed to convert $\displaystyle 19_{15}$ and $\displaystyle 25_{15}$ to base 10 to find their factors.

We have: .$\displaystyle \begin{Bmatrix}19_{15} &=& 24 &=& 2^3\cdot3\\ 25_{15} &=& 35 &=& 5\cdot7\end{Bmatrix}$

But I don't see how this helps.
We would need some divisibility theorems for base-15, wouldn't we?

For example, I can see that $\displaystyle (1)\;CA36D5$ is divisible by 5.
. . But is it divisible by 7?

3. Here is the result if you want to verify yours.
Spoiler:

fancymouse@fancymouse:~$bc bc 1.06.94 Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc. This is free software with ABSOLUTELY NO WARRANTY. For details type `warranty'. ibase=15 a=CA36D5 b=1E959B3 ibase=10 a%19 17 a%25 0 b%19 0 b%25 3 4. Of course, there's always my favorite method, brute force:$\displaystyle CA36D5_{15}=9630425$is divisible by 35 but not 24.$\displaystyle 1E959B3_{15}=22496568$is divisible by 24 but not 35. You could try long division in base 15. Or you could work with congruences:$\displaystyle 15^0\equiv{1}\equiv{1}\ (\text{mod }24)\displaystyle 15^1\equiv{15}*1\equiv{15}\ (\text{mod }24)\displaystyle 15^2\equiv{15}*15\equiv{9}\ (\text{mod }24)\$
and so on, then multiply each digit by the proper residue and take the sum.

Instead of 24 and 35, you could use 8, 3, 7, and 5 as moduli. That's probably what I'd use if I only had pencil and paper.
CA36D5 is divisible by 5 and not 3, so we only need to look at 7. It is the sum of digits, so 12+10+3+6+13+5 = 49, so it is divisible by 7 and therefore 35.
1E959B3 is divisible by 3 and not 5, so we only need to look at 8. It is an alternating sum, so 1-14+9-5+9-11+3 = -8, so it is divisible by 8 and therefore 24.

- Hollywood