If 68A3Y is divisible by 3, 7, and 11, find 4A+3Y?

Steps to Solve

1. Divisibility by 3, 7, and 11

If a number is divisible by 3, 7, and 11, it is also divisible by their least common multiple (LCM). Since 3, 7, and 11 are prime numbers, their LCM is simply their product.

2. Calculate the LCM

Calculate the LCM of 3, 7, and 11: $3 \times 7 \times 11 = 231$.

3. Divisibility by 231

The number 68A3Y must be divisible by 231. Therefore, $68A3Y = 231 \times k$, where $k$ is an integer.

4. Finding the range for k

$68030 \le 68A3Y \le 68939$ since A and Y are digits between 0 and 9. Divide the lower and upper bounds of the range by 231: $68030 / 231 \approx 294.5$ $68939 / 231 \approx 298.4$. Since $k$ must be an integer, the possible values for $k$ are 295, 296, 297, and 298.

5. Testing k values

   • If $k = 295$, $231 \times 295 = 68145$. Then $A = 1$ and $Y = 5$.
   • If $k = 296$, $231 \times 296 = 68376$. Then $A = 3$ and $Y = 6$.
   • If $k = 297$, $231 \times 297 = 68607$. Then $A = 6$ and $Y = 7$.
   • If $k = 298$, $231 \times 298 = 68838$. Then $A = 8$ and $Y = 8$.

6. Divisibility rule of 3

$6 + 8 + A + 3 + Y$ must be divisible by 3.

7. Checking divisibility by 3 for all possible A and Y values

   • If $A = 1$ and $Y = 5$, $6 + 8 + 1 + 3 + 5 = 23$, which is not divisible by 3.
   • If $A = 3$ and $Y = 6$, $6 + 8 + 3 + 3 + 6 = 26$, which is not divisible by 3.
   • If $A = 6$ and $Y = 7$, $6 + 8 + 6 + 3 + 7 = 30$, which is divisible by 3.
   • If $A = 8$ and $Y = 8$, $6 + 8 + 8 + 3 + 8 = 33$, which is divisible by 3.

8. Divisibility rule of 11 The alternating sum of digits must be divisible by 11. That is, $(6 - 8 + A - 3 + Y)$ must be divisible by 11.

9. Checking divisibility by 11 for A=6 and Y=7 $6 - 8 + 6 - 3 + 7 = 8$, which is not divisible by 11.

10. Checking divisibility by 11 for A=8 and Y=8 $6 - 8 + 8 - 3 + 8 = 11$, which is divisible by 11. So $A=8$ and $Y=8$.

11. Calculate 4A + 3Y

$4A + 3Y = (4 \times 8) + (3 \times 8) = 32 + 24 = 56$