Place the numbers 1 to 7 in the circles so that the sum of each row is 12. Use each number from 1 to 12 exactly once.

Place the numbers 1 to 7 in the circles so that the sum of each row is 12. Use each number from 1 through 12 exactly once. That way, one circle still has a sum of 17 and the others are made smaller by 2. Use each number from 1 to 12 exactly once. Mar 10, 2022 · You can solve the puzzle by moving the 2 into the same region as the 5. A suggested arrangement places 3 at the top, with 2, 1, and 7 to the right and 6, 5, and 4 to the left. Place the number 1 to 12 in the 12 circles so that the sum of the numbers in the six lines of the star is 26. Each straight line of three numbers will add up to 10. This video teaches you how to solve a number puzzle by using addition equations. Oct 6, 2023 · The problem of placing numbers 1-7 in circles so that the sum of each line equals 12 can be solved by arranging the numbers strategically. Jun 17, 2018 · Place the number 1 to 12 in the 12 circles so that the sum of the numbers in each of the six lines of the star is 26. Find more possible ways? Mar 26, 2024 · The sum of the odd digits is 1+3+5+7=16, so to fill out the top-right and bottom-left circles with a combined total of 14+14=28, we need to use both the 4 and the 8 from the top-left circle. Whether you're The sum of numbers from 1 to 7 is $$\frac {7 (7+1)} {2} = 28$$27(7+1) = 28. The numbers 1-7 are placed in each circle so that the total of every line is 12. Here are some other sums: Aug 8, 2023 · To solve the problem of placing the numbers 1 to 7 in circles so that the total of every line equals 12, we can start by understanding that we need to find a combination of these numbers that satisfies the condition. Can you arrange the numbers 1 to 7 in these regions so that for each of the lines, the sums of the numbers on either side are all the same? Jan 8, 2025 · You can place the numbers 1, 3, and 6 in the first circle, 2, 4, and 7 in the second circle, and 5 in the third circle. Sep 9, 2017 · Put the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 so the total of each row, column and diagonal add up to the same number [duplicate] Ask Question Asked 7 years, 11 months ago Modified 7 years, 11 months ago Mar 29, 2020 · It's not a spoiler to point out due to symmetry that for any valid solution, we can get 2x2 = 4 2 x 2 = 4 or 2x2x2 = 8 2 x 2 x 2 = 8 equivalent solutions by symmetry: flip L-R (swap the vertical columns), and flip each (or both) columns T-B. Dec 30, 2020 · Place the numbers from 1 to 12 in the circles without repetition to get a sum of 28, 29, 30 and 31. . Since the sum of each row is 12, the sum of the numbers in the outer circles must be $$12 \times 3 - C = 36 - C$$12×3−C = 36−C Apr 7, 2025 · The three lines in the figure below divide the circle into seven regions. ) Anyway, WLOG you can require 1 to be in the top-left Place the numbers 1 through 12 in the 12 circles so that the sum of the numbers in each of the six rows is 26. (If you want to keep the diagonals equal, you have to flip both columns, so only 2x2 = 4 2 x 2 = 4. fqqb jpt lpps tbem zrze amanqg xramcwe osndeq npd qmwpumqu