We can apply at max of 12 triggers in a table.

With close observation, you will note that each number in the list is in the middle of two prime numbers. Thus:

4 is in the middle of 3 and 5, 6 is in the middle of 5 and 7, 12 is in the middle of 11 and 13, 18 is in the middle of 17 and 19, 30 is in the middle of 29 and 31. 42 is in the middle of 41 and 43, 60 is in the middle of 59 and 61.

Therefore, the next number would be the one that is in the middle of the next two prime numbers, which is 72 (which is in the middle of 71 and 73).

Let y be the total number of loaves
a be the remaining loaves after 1st thief
b be the remaining loaves after 2nd thief
and c, d..so on after 3 and 4 th
e = 0 be the remaining loaves after 5th thief

a = y - (y/2 + 1/2)
= y/2 - 1/2

b = a - (a/2 + 1/2)
= a/2 - 1/2

c = b - (b/2 + 1/2)
= b/2 - 1/2

d = c - (c/2 + 1/2)
= c/2 - 1/2

e = d - (d/2 + 1/2)
= d/2 - 1/2
we know that, e = 0,
thus by substitution, we get
d = 1
c = 3
b = 7
a = 15
y = 31
Therefore total number of bread loaves in bakery was 31.

Suppose there are 9 balls

Let us give name to each ball B1 B2 B3 B4 B5 B6 B7 B8 B9

Now we will divide all the balls into 3 groups.

Group1 - B1 B2 B3

Group2 - B4 B5 B6

Group3 - B7 B8 B9

Step1 - Now weigh any two groups. Let's assume we choose Group1 on left side of the scale and Group2 on the right side.

So now when we weigh these two groups we can get 3 outcomes.

Weighing scale tilts on left - Group1 has a heavy ball.
Weighing scale tilts on right - Group2 has a heavy ball.
Weighing scale remains balanced - Group3 has a heavy ball.
Lets assume we got the outcome as 3. i.e Group 3 has a heavy ball.

Step2 - Now weigh any two balls from Group3. Lets assume we keep B7 on left side of the scale and B8 on right side.

So now when we weigh these two balls we can get 3 outcomes.

Weighing scale tilts on left - B7 is the heavy ball.
Weighing scale tilts on right - B8 is the heavy ball.
Weighing scale remains balanced - B9 is the heavy ball.
The conclusion we get from this Problem is that each time weigh. We element 2/3 of the balls.

As we came to conclusion that Group3 has the heavy ball from Step1, we remove 6 balls from the equation i.e (2/3) of 9.

Simillarly we do the ame thing for the Step2.

Now going with this conclusion. We have 6561 balls.

Step - 1

Divided into 3 groups

Group1 - 2187Balls

Group2 - 2187Balls

Group3 - 2187Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 2

Divided into 3 groups

Group1 - 729Balls

Group2 - 729Balls

Group3 - 729Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 3

Divided into 3 groups

Group1 - 243Balls

Group2 - 243Balls

Group3 - 243Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 4

Divided into 3 groups

Group1 - 81Balls

Group2 - 81Balls

Group3 - 81Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 5

Divided into 3 groups

Group1 - 27Balls

Group2 - 27Balls

Group3 - 27Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 6

Divided into 3 groups

Group1 - 9Balls

Group2 - 9Balls

Group3 - 9Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 7

Divided into 3 groups

Group1 - 3Balls

Group2 - 3Balls

Group3 - 3Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 8

So now when we weigh 2 balls out of 3 we can get 3 outcomes.

Weighing scale tilts on left - left side placed is the heavy ball.
Weighing scale tilts on right - right side placed is the heavy ball.
Weighing scale remains balanced - remaining ball is the heavy ball.
So the general answer to this question is, it is always multiple of 3 steps.

For 9 balls  32= 9. therefore 2 steps

For 6561 balls 38 = 6561 therefore 8 steps

There should be 90 instead of 80
so we can have a series
5, 5x3, 15x2, 30x3, 90x2, 180x3, 540x2 and 1080x3 and so on...

Let the number of apples in each tree of the 5 farmers be a, b, c, d,e respectively. Therefore total no of apples are 7a, 9b, 11c, 13d and 14e respectively.

Given,
7a+1 = 9b+3 = 11c-1 = 13d+3 = 14e-6 = x

9b+3 = 13d+3
===> 9b = 13d
so take b = 13 and d = 9

9b+3 = 9*13+3 = 120
ie, x=120

substituting
7a+1 = 120
7a = 119
a = 17

11c-1 = 120
11c = 121
c=11

14e-6 = 120
14e = 126
e = 9

Yields per tree in the orchards of d 3rd & 4th farmers= 11,9

7 boxes & 24 Balls.
Let, n = No. of boxes ; m = No. of fruits
3n+3 = m---------->(1)
4(n-1)= m---------->(2)
=>4n-4 = 3n+3;
=>4n-3n = 4+3
n = 7
Put n=7 in eqn(1)
=> 3(7)+3 = m
21+3 = m
m = 24;

Number of boxes = 7 and fruits = 24

Assume that the number is taken as XYZ.
Condition 1 => Z = sq.root(x) => Possible digits of X is 1,4,9
Condition 2 => Y = X + Z
X Y Z
1 2 1
4 6 2
9 12 3 => (wrong because it is 4 digit.)

and 121 is not divisible by 2

The answer is 462.