If the six-digit number 9x73x0 is divisible by 60, then the value of (3x +…

For the six-digit number 9x73x0 to be divisible by 60, it must be divisible by its co-prime factors: 3, 4, and 5. Since the number ends in 0, it is already divisible by 10 (which covers both 2 and 5). Therefore, we primarily need to check the divisibility rules for 3 and 4.

Divisibility by 4: A number is divisible by 4 if its last two digits form a number that is divisible by 4.
The last two digits are "x0". For a number ending in 0 to be divisible by 4 (like 00, 20, 40, 60, 80), the tens digit "x" must be an even number.
Possible values for x = 0, 2, 4, 6, 8.

Divisibility by 3:
A number is divisible by 3 if the sum of its digits is a multiple of 3.
Sum of digits = 9 + x + 7 + 3 + x + 0 = 19 + 2x.

Testing the possible even values for x:

If x = 0: Sum = 19 + 2(0) = 19 (Not divisible by 3)

If x = 2: Sum = 19 + 2(2) = 23 (Not divisible by 3)

If x = 4: Sum = 19 + 2(4) = 27 (Divisible by 3)

If x = 6: Sum = 19 + 2(6) = 31 (Not divisible by 3)

If x = 8: Sum = 19 + 2(8) = 35 (Not divisible by 3)

Therefore, the only valid digit that satisfies all conditions is x = 4.

Calculate the final expression:
Substitute x = 4 into the expression (3x + 2) / (3x - 2):
= [3(4) + 2] / [3(4) - 2]
= (12 + 2) / (12 - 2)
= 14 / 10
= 7/5