If the seven digit number $94 a 29 b 6$ is divisible by 72 , then what is the value of $(2 a+3 b)$ for $a \neq b$ ?

1. We have a seven digit number $94a29b6$ which is divisible by $72$

• Since $72 = 8 × 9$, the number must be divisible by both $8$ and $9$

2. For divisibility by $8$:

• Last three digits $9b6$ must be divisible by $8$
  • Try values for $b$ from $0$-$9$ where $9b6$ is divisible by $8$
  • $b = 3$ ($936 ÷ 8 = 117$)
  • $b = 7$ ($976 ÷ 8 = 122$)

3. For divisibility by $9$:

• Sum of digits must be divisible by $9$
  • $9 + 4 + a + 2 + 9 + b + 6 = 30 + a + b$ must be divisible by $9$

4. With $b = 3$:

• $30 + a + 3 = 33 + a$ must be divisible by $9$
  • When $a = 3$, sum $= 36$ ✓

but the question states $a$ and $b$ can't be equal. So:

5. With $b = 7$:

• $30 + a + 7 = 37 + a$ must be divisible by $9$
  • When $a = 8$, sum $= 45$ ✓

6. We need $a ≠ b$, so $a = 8$, $b = 7$ works.

7. Calculate $2a + 3b$:

= $2(8) + 3(7)$

= $16 + 21$

= $37$