Solve. Place the numbers 1, 2, 3, 4, 5, and 6 in the spaces so that each side will give the same sum.
Describe the strategy you used to find your solution(s).
See here.
How to Solve It?
Solve. Place the numbers 1, 2, 3, 4, 5, and 6 in the spaces so that each side will give the same sum.
Describe the strategy you used to find your solution(s).
See here.
1 3 5 horizontal
2 3 4 vertically. And vice versa
3 is common and sum is rest two is same both ways.
Thank you so much for your solution. However, a complete solution is already provided, which includes your one.
2+3+4=9. Therefore, wrong answer. The total should be 8.
If x1,x2,x3,x4 &x5 are 5 numbers ;
If Any xi comes in center i=1,2,3,4,5;
Then sum of two is sum of remaining two excluding xi;
If xi=1 remainings are 2,3,4 &5 :- 2+1+5( let it be Horizontal)=3+1+4(let it be vertical);
3+1+4 ( let it be Horizontal)=2+1+5(let it be vertical);
If xi=2 remainings are 1,3,4 & 5:- Solution not possible;
If xi=3 remainings are 1,2,4 &5 :- 1+3+5(H)=2+3+4(V);
2+3+4(H)=1+3+5(V);
If xi=4 remainings are 1,2,3 & 5:- Solution not possible;
If xi=5 remainings are 1,2,3,4 – (1+5+4)(H)=(2+5+3)(V);
(2+5+3)(H)=(1+5+4)
There are two possibilities in adding the five numbers to equal eight: 1,2,5 and 1,3,4. The number common to both permutations is 1. Therefore, the number in middle can only be 1.
Thank you
How did you solve problem number 10?