Question

I am neha,jee aspirant plz solve this question​

Answer 1

Answer:

1) 72 2) 144 3) 288

Step-by-step explanation:

Concept used---->

1) Number divisible by 9 if sum of its digit is divisible by 9 .

Given----> Digits A , B , and C are such that three digits A88 , 6B8 , 86C are divisible by 72 .

To find ----> Value of given determinant

Solution----> First we take first thre digit number which is A88 , it is divisible by 72 .

We know that , 72 = 9 × 8

So , given number is divisible by , 9 and 8 both ,

So, sum of digits of A88 is divisible by 9 .

A + 8 + 8 = 16 + A ,

Now A can take any values from 1 to 9 .

but A , can take value as 2 only because

16 + A = 16 + 2

= 18 , which is divisible by 9

If we take any other value sum is not divisible by 9 .

Now if we take value of A = 2 ,

First number = 288 , which is divisible by 8 also , so , 288 is divisible by 72 .

So , value of A = 2

Now , taking second number which is , 6B8 ,

Now sum of digits of second number = 6 + B + 8

= 14 + B

Now B can take only value which is equal to 4

14 + B = 14 + 4

= 18 , which is divisible by 9

So second number = 648 , which is also divisible by 8 .

So value of B = 4

Similarly , C = 4

Now A = 2 , B = 4 , C = 4

Now we put these values in determinant and solve the determinant ,

Value of determinat = 288

So determinant is divisible by 72 , 144 and 288 .

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