<s>[INST] Sum and Divisibility Puzzle I have $5$ positive integers: $a,b,c,d,e$. 
$a,b,c,d,e$ are all different, and $a\mid b\mid c\mid d\mid e$, in other words the ratios
$$\frac ba, \frac cb,   \frac dc,  \frac ed $$ are all integers.
$$a+b+c+d+e = 47.$$
I need to find out what $a,b,c,d,e$ are. Apparently there is only one solution to this.
I did some trial and error and came to an answer of $1,2,4,8,32$. But I really have no idea how to come to this conclusion more formally.
The problem comes from a section of my book that talks about prime factorization. 
I can figure that the direction that I need to go in is looking at the fact that 47 is prime and that adding the prime factorizations of a,b,c,d,e will give me 47 in some elegant way.
Can someone provide some direction?
 [/INST] First, if $a \geq 2$, then you'd have $b \geq (2)(2) = 4$, and $c\geq (2)(4) = 8$, and $d\geq (2)(8) = 16$ and $e\geq (2)(16) = 32$, since each of the "adjacent ratios" must be at least $2$.  But then $a + b + c + d + e \geq 62$, which doesn't work.  Hence $a = 1$.
Next, if $b = 3$, then $b, c, d, e$ are all divisible by $3$, which means $a + b + c + d + e$ is congruent to $1$ modulo $3$, which doesn't work since $47$ is congruent to $2$ rather than $1$.  Also, if $b \geq 4$, then $a+b+c+d+e > 47$ as above.  Therefore we must have $b = 2$, since $b > a = 1$.
Now $c = bk = 2k$ for some $k>1$.  If $c = (2)(3) = 6$, then $a+b+c+d+e = 0$ mod $3$, which again doesn't work, while if $c \geq (2)(4) = 8$, then $a+b+c+d+e > 47$ again.  Hence $c = (2)(2) = 4$.
At this point we have $(a, b, c) = (1, 2, 4)$, so $a+b+c=7$ and therefore $d+e = 40$.  Also, $d = 4m$ and $e = dn = 4mn$ for some integers $m, n > 1$.
Thus $40 = 4m + 4mn = 4(m + mn)$, so $10 = m + mn = m(1 +n)$.  The only way to write $10$ as the product of two integers both greater than $1$ is $10 = (2)(5)$, so either $m = 2$ and $1+n = 5$, or $m=5$ and $1+n = 2$.  The latter possibility is eliminated as it would mean $n=1$, so we conclude $m=2$ and $1+n=5$, so $n=4$.  That is, $d=8$ and $e=32$.
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