Brainsteem Quickfire Q6 [Win 2 SBD and 1 SBD]
Brainsteem Quickfire questions is a series of slightly easier, certainly faster, mathematical challenges in the Brainsteem family!
Winner will get 2 SBD! Second prize is 1 SBD to one person selected randomly from all other correct solutions!
The Question
Take the numbers 1 to 9 and place one number in each circle so that the sum of four numbers along each side of the triangle are all equal.
What is the largest possible such sum?
As is often the case, there is an insight that makes the answer much faster than randomly trying to add nine numbers.
Good luck!
Image is my own design and first appeared in giftedmathematics.com, my own blog
Prizes
To qualify you must upvote this post and write the answer in a comment.
To win First Prize you must include the answer and a valid method.
The *First Prize will be 2 SBD.
The Second Prize will be 1 SBD awarded to any other correct answer in the comments section, chosen at random.
Also remember that your upvote generates curation rewards for you and that 25% of all earnings are given to upvoters.
Prizes will be awarded after the 7 days have expired. Winners will be announced in the comments below after the draw.
The Prizes will increase with more participation and as the earnings increase.
The main thing is to enjoy the challenge and maybe learn something new.
Come and check out Brainsteem challenges and other Mathematical posts currently live!
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sum of side = 23
solution: sum of 1-9 = 45
sum of three sides = 45 + sum of corners = sum of side x 3
sum of corners = (sum of side x 3) -45
thus the sum of corners is divisible by 3 , we now can try out with limited possibility
oh I forgot to include the number on each side lol
To maximize the sum, start by putting 9, 8 and 7 in the corners so they can each be used on two sides. That leaves sides of sums 15, 16, and 17, each with two blanks to fill in. The sum of the remaining numbers is 21, and we need to split them into three pairs with sums that are three consecutive numbers.
21 as the sum of three consecutive numbers is 6+7+8. There are multiple ways of splitting the numbers 1-6 into three pairs whose sums are 6, 7, and 8. One such way is 2+4, 1+6, and 3+5. The resulting triangle is
Basically, match up the pair totaling 6 with the pair totaling 17. Do the same for 7 with 16, and 8 with 15. The maximum possible sum on each side of the triangle is 23.
Ok to get the maximum amount, start offcourse with the highest number clockwise (9,8,7) in the corners, then counterclockwise, put the next 3 in (6,5,4) and again counterclockwise but start 1 row later (3,2,1). With that you get top row (9,2,4,8) rightrow (8,5,3,7) and left row (9,1,6,7) and all equal 23.
The 3 corners get counted twice so they should have the highest numbers: 9, 8, 7
9 should get the 2 smallest numbers in it's 2 sides: 1, 2
We now have 9+1=10 & 9+2=11 add 7 and 8, must be 17 and 19 (18 and 18 won't work)
The numbers left are 3, 4, 5, 6
Either 5 and 3 or 6 and 4 go with the 17 and 19 = either 22 ( 5 and 3) or 23 (6 and 4)
The remaining side has 7 and 8 for 15. Add the 2 remaining number and get either 15 + 8 = 23
or 15 + 10 = 25
So the total that matches is 23 as the highest possible equal total.
9 1 6 7
,2 , , 5
,,4 3
,,,8