A seven-digit number 489y5z6 is divisible by 72. Which of the options gives the highest possible product of y and z?

To find the values of y and z such that the seven-digit number 489y5z6 is divisible by 72, we need to check divisibility by both 8 and 9, as 72 = 8 x 9. Divisibility by 8: The last three digits, 526, must be divisible by 8. Let's test possible values for z: For z = 0: 506 (not divisible by 8) For z = 1: 516 (not divisible by 8) For z = 2: 526 (not divisible by 8) For z = 3: 536 (divisible by 8) For z = 4: 546 (not divisible by 8) For z = 5: 556 (not divisible by 8) For z = 6: 566 (not divisible by 8) For z = 7:576 (divisible by 8) For z = 8: 586 (not divisible by 8) For z = 9: 596 (not divisible by 8) so take the highest value of  z =7 Divisibility by 9: The sum of the digits, 4+8+9+y+5+z+6=32+y+z, must be divisible by 9. Let's test possible values of z and find corresponding y: z =7 32+7+y=39+y y =6 , =39+6=45 is divisible by 9. now product of y and z =7×6 =42.