Pascal's triangle/Puzzle: Difference between revisions

</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]