Playing With Numbers

A = 8

Step 1: Look at the units column: A + 3 must give a number whose units digit is 1. Step 2: A + 3 = 1 → A = −2 (not valid). A + 3 = 11 → A = 8. A + 3 = 21 → A = 18 (not a digit). So A = 8. Step 3: Verify tens column: 1 + A = 1 + 8 = 9, but we need 0 at tens place with a carry. 1 + 8 = 9 (no carry needed from units since 8+3=11 gives carry 1). So tens: 1 + 1 + 8 = 10, write 0 carry 1. Step 4: Hundreds: 1 + 3 + 1 = 5. ✓ So 318 + 183 = 501. Confirmed A = 8.

A = 6, B = 1

Step 1: Look at units column: 9 + A must end in digit 5. Step 2: 9 + A = 5 → A = −4 (invalid). 9 + A = 15 → A = 6. So A = 6, and 1 is carried over. Step 3: Tens column: 1 (carry) + B + 4 = 6 → B + 5 = 6 → B = 1. Step 4: Verify: 19 + 46 = 65. ✓ A common mistake is forgetting to add the carry of 1 in the tens column, which leads to B = 2 incorrectly.

45387

Step 1: A number is divisible by 9 if the sum of its digits is divisible by 9. Step 2: Check 45387: 4 + 5 + 3 + 8 + 7 = 27. Is 27 divisible by 9? 27 ÷ 9 = 3. Yes! Step 3: Check 3518: 3 + 5 + 1 + 8 = 17. 17 is not divisible by 9. Step 4: Check 7241: 7 + 2 + 4 + 1 = 14. Not divisible by 9. Check 8053: 8 + 0 + 5 + 3 = 16. Not divisible by 9. So only 45387 is divisible by 9.

x = 0, 3, 6 or 9

Step 1: A number is divisible by 3 if the sum of its digits is divisible by 3. Step 2: Sum of known digits = 3 + 1 + 5 = 9. Total sum = 9 + x. Step 3: For 9 + x to be divisible by 3, x must be such that 9 + x is a multiple of 3. Since 9 is already a multiple of 3, x itself must be a multiple of 3. Step 4: Digits that are multiples of 3: x = 0, 3, 6, 9. A common mistake is thinking only x = 3 works, but all multiples of 3 among digits are valid answers.