Brainsteem Mathematics Challenges: Integer Quotients
Haven't posted a mathematical puzzle for a while; time to remedy that!
A fairly gentle start, inspired by a Junior Challenge.
Question
The aim is to fill the inverted pyramid of squares with six different positive integers, one in each square.
Each square below the first row is the quotient of the two squares above it, reading from left to right - see the smaller diagram. The numbers in the first row are, obviously, a free choice.
. 
What is the smallest possible sum of the six numbers?
images: @rycharde, created with Geogebra
The first correct answer and further interesting comments will be rewarded with an upvote.
Enjoy!
More Brainsteems coming soon...
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@rycharde - no doubt wrong, but, 4-2-4 /2-2/1 = 15?
i think i missunderstoof the question :P
The question asks for six different positive integers :-) - otherwise six 1s would solve it!
Also, the division is always left-to-right.
Just brute-forcing off the top of my head... I get 99.
2 on the bottom
6 and 3 in the middle row
72, 12, and 4 on the top row
Hi, that's a fair first iteration, but there is a smaller solution.
Thanks!
Of course... smacks self
I should've seen this pair of solutions from the get-go, both totaling 71...
3 on the bottom
6 and 2 in the middle row
48, 8, and 4 on the top row
OR
4 on the bottom
8 and 2 in the middle row
48, 6, and 3 on the top row
Yes :-)
There was another puzzle that I shall repost as nobody solved it - similar solution where the obvious minimum is not the real one! I kinda like such minimax problems.
Devised this problem myself as the one in the maths challenge paper was far too easy. Tested this out with a small group of gifted kids at school - mainly as training in how to devise problems, not just solve them.
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