Question $1 .$
In the following magic triangle, arrange the numbers from 1 to 6, so that you get the same sum on all its sides.
Solution:
Step 1: Complete the corners with smaller numbers 1,2 and $3 .$
Step 2: The side having smallest numbers 1 \& 2 are to be filled with the greatest number 6 , the second smallest 1 \& 3 side to be filled with the second largest 5 at the middle and so on.
The magic sum is $1+6+2=2+4+3=3+5+1=9 .$ Some other ways are given below.
The magic sum $=1+6+3=3+2+5=5+4+1=10$.
The magic sum $6+1+4=4+5+2=2+3+6=11$.
The magic sum $4+3+5=5+1+6=6+2+4=12$.
Question $2 .$
Using the numbers from 1 to 9
(i) Can you form a magic triangle?
(ii) How many magic triangles can be formed?
(iii) What are the sums of the sides of the magic triangle?
(i) Yes, we can form
(ii) 5
(iii)
Sums are 17, 19, 20, 21 and 23.
Question $3 .$
Arrange the odd numbers from 1 to 17 without repetition to get a sum of 30 on each side of the magic triangle.
Solution:
The odd numbers between 1 to 17 are $1,3,5,7,9,11,13,15,17$.
Step 1: Place the smaller numbers $1,3,5$ on the comers.
Step 2: Arrange another set of smaller numbers 7,9 and 11 on each side.
Step 3: Arrange the remaining numbers 13,15,17 to give the total 30
Magic sum = 30.
Question $4 .$
Put the numbers $1,2,3,4,5,6$ and 7 in the circles so that each straight line of three numbers add up to the same total.
Solution:
Here there is even number of terms. Also we know that $1+6=7,2+5=7,3+4=7 ;$ so placing 7 at the centre, and the pairs $(1,6)(2,5)$ and $(3,4)$ at. the opposite ends we get, the answer.
Question $5 .$
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 . Use each number from 1 to 12 exactly once. Find more possible ways.
Solution:
The given star can be viewed as two magical triangular as.
Now the required arrangement is
Some other arrangements are