Alpha Numeric: Level 1

Solution (B): Look at the units column: \(A + A = 8\). This means \(2A = 8\), so \(A = 4\). Check the tens column: \(3 + 5 = 8\). The problem is \(34 + 54 = 88\). This is correct.

Solution (A): Look at the units column: \(5 + A = 0\). This must be 10, so we carry a 1. If \(5 + A = 10\), then \(A = 5\). Check the tens column: \((A + 3) + 1(\text{carry}) = 9\). \((5 + 3) + 1 = 8 + 1 = 9\). This is correct. The problem is \(55 + 35 = 90\).

Solution (C): This problem can be written as \((10A + A) + (10B + B) = 66\). \(11A + 11B = 66\). Factor out 11: \(11(A + B) = 66\). Divide by 11: \(A + B = 6\).

Solution (E): For a number to be divisible by 3, the sum of its digits must be a multiple of 3. Sum = \(1 + 2 + X + 3 = 6 + X\). We need \(6+X\) to be a multiple of 3. Since 6 is already a multiple of 3, X must also be a multiple of 3. Possible values for X are {0, 3, 6, 9}. Of the options given, only 9 is in this set.

Solution (B): Look at the units column. We have \(0 – A = 2\). This means we must borrow from the tens place (A). \(10 – A = 2\), which means \(A = 8\). Check the tens column: We borrowed 1 from A, so it becomes \(A-1\). \((A-1) – 4 = 3\). Substitute \(A=8\): \((8-1) – 4 = 7 – 4 = 3\). This is correct. The problem is \(80 – 48 = 32\).

Solution (C): We are looking for a digit A such that \(A \times 4\) results in a two-digit number that ends in A. Let’s test the options: (A) \(4 \times 4 = 16\). This ends in 6, not 4. (B) \(5 \times 4 = 20\). This ends in 0, not 5. (C) \(6 \times 4 = 24\). This ends in 4, not 6. *Correction:* Let’s re-read the problem. \(2A\) is a two-digit number. The tens digit is 2, and the units digit is A. Let’s test again: (A) \(4 \times 4 = 16\). Does 16 = 2A = 24? No. (B) \(5 \times 4 = 20\). Does 20 = 2A = 25? No. (C) \(6 \times 4 = 24\). Does 24 = 2A = 26? No. (D) \(7 \times 4 = 28\). Does 28 = 2A = 27? No. (E) \(8 \times 4 = 32\). Does 32 = 2A = 28? No. There is an error in my question creation. Let’s fix Q6. `A \times 6 = 3A`. Test: (A) 4: \(4 \times 6 = 24\). Does 24 = 34? No. Test: (B) 5: \(5 \times 6 = 30\). Does 30 = 35? No. Test: (C) 6: \(6 \times 6 = 36\). Does 36 = 36? Yes.

Solution (C): We are looking for a digit A such that \(A \times 6\) results in a two-digit number that has a tens digit of 3 and a units digit of A. Let’s test the options: (A) \(4 \times 6 = 24\). This does not equal 34. (B) \(5 \times 6 = 30\). This does not equal 35. (C) \(6 \times 6 = 36\). This equals 3A (where A=6). This is correct. (D) \(7 \times 6 = 42\). This does not equal 37. (E) \(8 \times 6 = 48\). This does not equal 38.

Solution (B): We need two numbers that multiply to 21 and add to 10. The factor pairs of 21 are (1, 21) and (3, 7). \(1 + 21 = 22\). (Incorrect) \(3 + 7 = 10\). (Correct) The two integers are 3 and 7. The smaller integer is 3.

Solution (D): This is an algebra problem. `AA` is \(10A + A = 11A\). \(11A + 55 = 77\). \(11A = 77 – 55\). \(11A = 22\). \(A = 2\). Check: \(22 + 55 = 77\).

Solution (A): Look at the tens column: \(B + 6\) results in \(CA\). Since B and 6 are single digits, their sum must be less than 20 (e.g., \(9+6=15\)). This means the carry-over digit \(C\) must be 1. (We can solve the rest: Units: \(A+B=B \implies A=0\). Tens: \(B+6 = 1A = 10 \implies B=4\). Check: \(40 + 64 = 104\). This works. \(C=1\).)

Solution (C): For a number to be divisible by 3, the sum of its digits must be a multiple of 3. Sum = \(5 + X + 1 = 6 + X\). We need \(6+X\) to be a multiple of 3. Since 6 is already a multiple of 3, X must also be a multiple of 3. Possible values for X are {0, 3, 6, 9}. Of the options given, only 3 is in this set.