Seven Sportsmen

The number of middle batsman can be 2 and the series can be

3 6 4 2 9 8 1

Here,
product of first three = 3 * 6 * 4 = 72
product of middle three = 4 * 2 * 9 = 72
product of last three = 9 * 8 * 1 = 72

-Pradeep

2 is the correct answer...did u came to the answer by hit and trial or do u have a logic 4 it...for the latter case , please explain your logic ...

The answer has a logic behind it.

The single digit numbers are 0 to 9. But there are only 7 batsman and each have different digits so we need to first ignore 3 digits from 0 to 9 and then use rest of the numbers to create the sequence as per given condition.


Ignoring the Digits

The three digits to be ignored are 0, 5, 7.

0: If this number is used then it can be shared between maximum two groups out of left, middle, and right. And any number multiplied by 0 is 0 so the multiplication of digits of those two groups will be 0 and the multiplication of digits of third group will be non-zero which will not satisfy the given condition. So we need to ignore this.

5,7: If any of these number is used then it can be shared between maximum two groups out of left, middle, and right. These are prime numbers and there is no other digit between 1 to 9 which is multiple of these digits. So if these are used there will no digit in the third group to balance the effect of these digit and the multiplication of third group can't be equal to first and second group. There are other prime numbers 2 and 3. But there ate other digits in list (4, 6, 8, 9) which have 2 and 3 ans their factors and can balance the effect of using 2 and 3.


Determining the Sequence

We have number 1, 2, 3, 4, 6, 8, 9 to to create sequence as per given condition.

To determine the sequence lets find the prime factors of above numbers:
2 = 2
3 = 3
4 = 2 * 2
6 = 2 * 3
8 = 2 * 2 * 2
9 = 3 * 3

So we have seven 2's and four 3's.

Now lets see the cases when we have any other digit instead of 2 in the middle:

1: If 1 is the middle digit then we have seven 2's to be distributed among left and right group. As the count is not even so it will not balance the left and right group. So we can't ahve 1 as middle digit.

3,6,9: If we have 3 or 6 in middle, we are left with three 3's in other digit which can't balance number of 3's in left and right group. If we have 9 in middle then we are left with only two 3's in other digits which can only balance one group with middle group. So 3, 6, and 9 are not suitable for middle digit.

4,8: If we have 4 in middle then we are left with five 2's in other digits which can't balance number of 2's in left and right group. If we have 8 in middle then we are left with only four 2's in other digits which are not sufficient to balance the other two group with middle group. So 4 and 8 are not suitable for middle digit.

Now we are left with 2 as the middle digit. Let's see why two works.

When we have 2 as middle digit we remain with six 2's and four 3's which can be evenly distributed among left and right group and can balance the equation.

When the 2's and 3's are evenly distributed we have three 2's and two 3's on each group making 2 * 2 * 2 * 3 * 3 = 72.

We already have 2 as middle digit so for middle group to make 71 the other two digit of middle group should multiply to make 36 and the only possible answer is 4 * 9. So the other two digit of middle group are 4 and 9.

Lets consider 4 is shared with left group and 9 is shared with right group is the sequence for middle group is "4, 2, 9".

Now for the left group to make 72 the other two digit should multiply to make 18 and with the digit 1, 3, 6, and 8 left the only possible answer is 3 and 8.

And the digit 8 and 1 become part of right group with 9 and make 72 as the multiplication. And thus the sequence is

3 6 4 2 9 8 1

PS: We can interchange the position of 3 with 6 and 8 with 1, which can give more possible sequence. But need to have 2 as middle digit.


Let me know if any item need more clarification.

-Pradeep