Alpha Numeric: Level 2

Solution (E): Units column: \(B + A = 4\) or \(B + A = 14\). Tens column: \(A + B\) or \((A + B + 1)\) must equal 15. If \(A+B = 4\), the tens column (with carry) would be 4 or 5, not 15. Therefore, \(A + B\) must be 14. This makes the units column \(B+A = 14\) (carry a 1). Tens column: \((A + B) + 1 = 14 + 1 = 15\). This matches. We need two distinct digits A and B that sum to 14. Possible pairs: (5, 9) and (6, 8). Possible values for A are {5, 6, 8, 9}. Of the options, 8 is a possible value.

Solution (C): For a number to be divisible by 9, the sum of its digits must be a multiple of 9. Sum = \(4 + X + 2 + X = 6 + 2X\). We test values for X to see which makes \(6 + 2X\) a multiple of 9. If \(X=6\), Sum = \(6 + 2(6) = 6 + 12 = 18\). 18 is a multiple of 9. This is the only single digit that works.

Solution (E): Units: \(B \times 3\) must end in the digit B. The only non-zero digit for this to work is \(B=5\) (\(5 \times 3 = 15\)). We carry a 1. Tens: \((A \times 3) + 1\) must end in 9. This means \(A \times 3\) must end in 8. The only digit A for which this works is \(A=6\) (\(6 \times 3 = 18\)). Check: \(A=6, B=5\). The problem is \(65 \times 3\). \(65 \times 3 = 195\). This fits the pattern `C9B` where \(C=1, B=5\). A=6, B=5, C=1 are distinct. \(A+B = 6+5 = 11\).

Solution (D): Units column: \(4 + B + A = 2\) or \(12\) or \(22\). Since A, B are positive, it must be 12 or 22. Tens column: \(A + 3 + 2 + (\text{carry}) = 13\). Case 1: Units sum is 12. \(A+B+4=12 \implies A+B=8\). Carry is 1. Tens: \((A + 3 + 2) + 1 = 13 \implies A+6 = 13 \implies A = 7\). If \(A=7\), then \(7+B=8 \implies B=1\). Check: \(74 + 31 + 27 = 105 + 27 = 132\). This is correct.

Solution (C): 1. From \(A+A = 1B\), \(2A\) is a two-digit number starting with 1. This means \(A\) can be {5, 6, 7, 8, 9}. (B=0, 2, 4, 6, 8 respectively). 2. We are told B is non-zero, so \(A=5, B=0\) is out. 3. Test the remaining pairs with \(B+C = A\). – If \(A=6, B=2\): \(2+C=6 \implies C=4\). \(A+C = 6+4=10\). – If \(A=7, B=4\): \(4+C=7 \implies C=3\). \(A+C = 7+3=10\). – If \(A=8, B=6\): \(6+C=8 \implies C=2\). \(A+C = 8+2=10\). – If \(A=9, B=8\): \(8+C=9 \implies C=1\). \(A+C = 9+1=10\). In all possible cases, \(A+C\) must be 10.

Solution (C): The divisibility rule for 11 is that the alternating sum of digits is a multiple of 11. Sum 1: \(5 + 7 = 12\) Sum 2: \(X + X = 2X\) Difference: \(12 – 2X\) must be 0, 11, -11, etc. If \(12 – 2X = 0\), then \(2X = 12\), so \(X = 6\). If \(12 – 2X = 11\), then \(2X = 1\), so \(X = 0.5\) (not a digit). The only solution is X=6.

Solution (C): We solve from right to left (units to hundreds). Units: \(B – 7 = 5\). We must have borrowed. \(12 – 7 = 5 \implies B = 2\). Tens: We borrowed from 3, so it’s 2. \(2 – C = 4\). We must have borrowed. \(12 – C = 4 \implies C = 8\). Hundreds: We borrowed from A. \((A-1) – 1 = 5 \implies A-2 = 5 \implies A = 7\). Solution: A=7, B=2, C=8. They are distinct. Sum: \(7 + 2 + 8 = 17\).

Solution (D): Units: \(A \times B\) must end in B. This happens if B=0, B=5, or A=1, or A=6. Case 1: B=5. (A must be odd). If \(A=1\): \(11 \times 5 = 55\). This fits `CBB` if \(C=0\). A=1, B=5, C=0. These are distinct. Let’s check A=3: \(33 \times 5 = 165\). `CBB` is 165. B must be 6. But we assumed B=5. Contradiction. The only solution is A=1, B=5, C=0. Sum: \(1 + 5 + 0 = 6\).

Solution (E): 1. For `1,28X` (div by 3): Sum = \(1+2+8+X = 11+X\). \(11+X\) must be a multiple of 3. Possible X: \(X=1\) (sum=12), \(X=4\) (sum=15), \(X=7\) (sum=18). 2. For `5,43Y` (div by 4): The last two digits `3Y` must be a multiple of 4. Possible `3Y`: 32 (Y=2) or 36 (Y=6). Possible Y: {2, 6}. 3. Possible sums \(X+Y\): \(1+2=3\), \(1+6=7\), \(4+2=6\), \(4+6=10\), \(7+2=9\), \(7+6=13\). The set of possible sums is {3, 6, 7, 9, 10, 13}. Of the options, 13 is in this set.

Solution (D): This is a classic problem. We can assume, without loss of generality, that \(A < B < C\). The only set of distinct positive integers that satisfies this is {1, 2, 3}. Check: Product: \(1 \times 2 \times 3 = 6\). Sum: \(1 + 2 + 3 = 6\). The sum \(A+B+C\) is 6.