Pascal's triangle/Puzzle: Difference between revisions
Content added Content deleted
(→{{header|jq}}: Algebraic solution) |
|||
Line 1,740: | Line 1,740: | ||
<pre> |
<pre> |
||
[5,13,8] |
[5,13,8] |
||
</pre> |
|||
===Algebraic solution=== |
|||
As noted elsewhere on this page, elementary considerations show that |
|||
the apex (top) value and the value in the third row (mid) can be written as: |
|||
<pre> |
|||
top = 4(a+b) + 7(x+z) |
|||
mid = 2x + 2a + z |
|||
</pre> |
|||
where a and b are the known values at the base. |
|||
Plugging in the known values yields: |
|||
<pre> |
|||
151 = 4 * 15 + 7(x+z) |
|||
40 = 2x + 22 + z |
|||
</pre> |
|||
i.e. |
|||
<pre> |
|||
13 = x +z |
|||
18 = 2x+z |
|||
</pre> |
|||
These two occasions can solved trivially, but for fun, let's use the linear equation solver |
|||
at https://rosettacode.org/wiki/Cramer%27s_rule#jq |
|||
<syntaxhighlight lang=jq> |
|||
include "rc-cramers-rule"; |
|||
cramer([[1,1], [2,1]] ; [13, 18]) |
|||
</syntaxhighlight> |
|||
This gives the solution for [x,z] as: |
|||
{{output}} |
|||
<pre> |
|||
[5,8] |
|||
</pre> |
</pre> |
||