Solve. (a) Using each of the digits 0, 1, 2, 3, 4, 5, 6 and 7 once, fill in the spaces below so that the difference is 8 in the hundreds place, 8 in the tens place and 9 in the ones place.
(b) Is there more than one solution to this problem? How?
(c) Is there more than one solution in which the digit in the thousands place of the difference is 0?
See here. A solution is given below.
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?