A six-digit number 3 5 a 7 2 b is divisible by 72. How many possible values does the number have?

72 = $8\times 9$. For 8, the last three digits $72b$ must be a multiple of 8: 720 and 728 work, so $b=0$ or 8. For 9, digit sum $17+a+b$ must be a multiple of 9: $b=0$ → $a=1$; $b=8$ → $a=2$. Two numbers: 351720 and 352728.