Alpha Numeric: Level 3

Solution (D): 1. Units: \(C+A = 2\) or \(C+A = 12\). 2. Hundreds: \(A+C + (\text{carry from tens}) = 13\). 3. Since \(A+C\) must be 12 (or 11), \(C+A=2\) is impossible (as A, C must be non-zero for 3-digit numbers). 4. So, Units: \(C+A = 12\). (Carry a 1). 5. Tens: \(B+B + 1 = 3\) or \(B+B+1 = 13\). 6. If \(2B+1=3\), \(2B=2 \implies B=1\). (No carry to hundreds). Hundreds would be \(A+C+0 = 12\). This is consistent. 7. If \(2B+1=13\), \(2B=12 \implies B=6\). (Carry 1 to hundreds). Hundreds would be \(A+C+1 = 13 \implies A+C=12\). This is also consistent. 8. We must check “distinct digits”. If B=1, A+C=12. Possible (A,C) pairs: (3,9), (4,8), (5,7). All are distinct from B=1. If B=6, A+C=12. Possible (A,C) pairs: (3,9), (4,8), (5,7). The digits must be distinct from B=6. All pairs are valid. *Correction:* Let’s re-read the image. Q1 `XYZ + ZYX = 848`. My question is different. Ah, I see `A+A = CB` and `C+B=5` (Q7). Let’s use that logic. If \(B=1\), A,C can be (3,9). e.g., 319+913 = 1232. Not 1332. If \(B=6\), A,C can be (3,9). e.g., 369+963 = 1332. This works. So B must be 6.

Solution (B): 1. From `A+A = CB`, \(2A\) is a 2-digit number `10C + B`. 2. Since A is a digit, the largest `2A` can be is 18 (if A=9). This means the tens digit, C, must be 1. 3. From `C+B = 5`, if \(C=1\), then \(1+B=5 \implies B=4\). 4. Now use `2A = 10C + B`. 5. \(2A = 10(1) + 4 \implies 2A = 14 \implies A = 7\). 6. The digits are A=7, B=4, C=1. They are distinct and non-zero. This is the solution.

Solution (A): 1. `AA` is \(11 \times A\). The equation is \((11 \times A) \times A = 39A\). 2. \(11 \times A^2 = 390 + A\). 3. We can test the options. (A) \(A=6\): \(11 \times (6^2) = 11 \times 36 = 396\). \(390+A = 390+6 = 396\). This matches. (B) \(A=5\): \(11 \times (5^2) = 11 \times 25 = 275\). \(390+5 = 395\). (No). (C) \(A=7\): \(11 \times (7^2) = 11 \times 49 = 539\). \(390+7 = 397\). (No).

Solution (C): 1. Units: \(Z+X = 8\) or \(Z+X = 18\). 2. Tens: \(Y+Y + (\text{carry}) = 9\) or \(19\). 3. Hundreds: \(X+Z + (\text{carry}) = 9\). 4. From Tens: \(2Y + \text{carry}\) must be odd. Since \(2Y\) is even, the carry from the units *must be 1*. 5. This means from Units: \(Z+X = 18\). (The only pair is 9,9, but they must be distinct. This is a contradiction). *Rethink:* Let’s re-check Tens. \(2Y + \text{carry} = 9\) or \(19\). *Case 1:* Carry from units = 0. Then \(Z+X = 8\). Tens: \(2Y+0 = 9\) (impossible) or \(2Y+0 = 19\) (impossible). *Case 2:* Carry from units = 1. Then \(Z+X = 18\). (So X=9, Z=9). But X and Z must be distinct. *Rethink again:* 1. Units: \(Z+X=8\) (no carry) or \(Z+X=18\) (carry 1). 2. Tens: \(Y+Y = 9\) (no carry) or \(Y+Y=19\) (no carry) or \(Y+Y+1=9\) (carry 1) or \(Y+Y+1=19\) (carry 1). 3. `2Y=9` (no), `2Y=19` (no). 4. `2Y+1=9 \implies 2Y=8 \implies Y=4\). (This implies carry from units=1). 5. `2Y+1=19 \implies 2Y=18 \implies Y=9\). (This implies carry from units=1). This is complex. Let’s restart. Units: \(Z+X = 8\) (carry 0) or \(Z+X = 18\) (carry 1). Tens: \(Y+Y = 9\) (carry 0) or `… = 19` (carry 1) … `2Y + \text{carry}_1 = 9 \text{ or } 19`. Hundreds: \(X+Z + \text{carry}_2 = 9\). From Tens: \(2Y + \text{carry}_1\) must be 9 or 19. If `carry_1 = 0` (from units): `2Y=9` (no) or `2Y=19` (no). If `carry_1 = 1` (from units): `2Y+1=9 \implies Y=4` (carry_2=0) OR `2Y+1=19 \implies Y=9` (carry_2=1). So, we must have `carry_1 = 1`, which means \(Z+X=18\). This means \(Z=9, X=9\). But they must be distinct. This implies my analysis of Q5 in the L3 brainstorming was wrong. `XYZ + ZYX = 998`. Units: `X+Z=8` or 18. Tens: `Y+Y = 9` (if no carry) or `Y+Y+1 = 9` (if carry). OR `Y+Y=19` or `Y+Y+1=19`. `2Y=9` (no). `2Y=19` (no). `2Y+1=9 \implies 2Y=8 \implies Y=4`. This means carry from units=1. `2Y+1=19 \implies 2Y=18 \implies Y=9`. This means carry from units=1. *Contradiction!* If Y=4, carry from units must be 1. So `X+Z=18`. `X=9, Z=9`. Not distinct. If Y=9, carry from units must be 1. So `X+Z=18`. `X=9, Z=9`. Not distinct. This question has no solution with distinct digits. Let’s re-use Q1 from the L3 brainstorming. `ABC + CBA = 1332`. Units: C+A = 12 (carry 1) Tens: B+B+1 = 3 (carry 0) or B+B+1 = 13 (carry 1) Hundreds: A+C + (carry) = 13 Case 1: `2B+1=3 \implies B=1`. Carry=0. Hundreds: `A+C+0=13`. Contradicts `A+C=12`. Case 2: `2B+1=13 \implies B=6`. Carry=1. Hundreds: `A+C+1=13 \implies A+C=12`. This is consistent. Solution is `B=6`, `A+C=12`. Question: What is `A+B+C`? `A+B+C = (A+C) + B = 12 + 6 = 18`.

Solution (D): 1. Units: \(C+A = 12\). (Must be 12, not 2, to get 1332. Carry 1). 2. Tens: \(B+B+1 = 3\) or \(B+B+1 = 13\). 3. Hundreds: \(A+C + (\text{carry from tens}) = 13\). 4. Case 1: Tens sum to 3. \(2B+1=3 \implies 2B=2 \implies B=1\). (Carry 0). Hundreds: \(A+C+0 = 13 \implies A+C=13\). This contradicts \(A+C=12\). 5. Case 2: Tens sum to 13. \(2B+1=13 \implies 2B=12 \implies B=6\). (Carry 1). Hundreds: \(A+C+1 = 13 \implies A+C=12\). This matches the units column. 6. Solution: \(B=6\) and \(A+C=12\). The digits must be distinct. e.g., A=4, C=8, B=6. (468 + 864 = 1332). 7. Sum: \(A+B+C = (A+C) + B = 12 + 6 = 18\).

Solution (D): 1. Divisible by 4: The last two digits, `B2`, must be a multiple of 4. Possible `B2`: 12, 32, 52, 72, 92. So \(B\) can be {1, 3, 5, 7, 9}. 2. Divisible by 9: Sum of digits must be a multiple of 9. Sum = \(7+A+5+B+2 = 14+A+B\). 3. Test cases for B: – If \(B=1\), Sum = \(15+A\). \(A=3\) (sum 18). \(A+B = 4\). – If \(B=3\), Sum = \(17+A\). \(A=1\) (sum 18). \(A+B = 4\). – If \(B=5\), Sum = \(19+A\). \(A=8\) (sum 27). \(A+B = 13\). – If \(B=7\), Sum = \(21+A\). \(A=6\) (sum 27). \(A+B = 13\). – If \(B=9\), Sum = \(23+A\). \(A=4\) (sum 27). \(A+B = 13\). The largest possible value for \(A+B\) is 13.

Solution (A): 1. `AA * B = (11*A) * B = 11AB`. 2. `CBC = 101C + 10B`. 3. `11AB = 101C + 10B`. 4. `11AB – 10B = 101C \implies B(11A – 10) = 101C`. 5. Since 101 is prime, one of the terms on the left must be a multiple of 101. 6. `B` is a single digit, so it can’t be. 7. `(11A – 10)` must be a multiple of 101. 8. Test A: If \(A=1\), \(11-10=1\). If \(A=9\), \(11(9)-10 = 99-10 = 89\). *Rethink:* My logic `B(11A-10) = 101C` is correct. If `A=9`, `B(89) = 101C`. Since 89 and 101 are prime, this can’t be true unless B=101 or A=101. *Rethink:* Let’s test `A=9`. `11 \times 9 \times B = 99B`. `99B = 101C + 10B \implies 89B = 101C`. Since 89 and 101 are prime, this only holds if `B=101` and `C=89`. Not possible. Let’s re-use the *other* puzzle I made. `AA \times AA = ABA` (from image). \(A=1 \implies 11 \times 11 = 121\). This fits `ABA`! \(A=1, B=2\). Question: A+B? Answer: 3.

Solution (C): This is from image 8. We test values for A. If \(A=1\): `11 x 11 = 121`. This fits the pattern `ABA` where \(A=1\) and \(B=2\). A and B are distinct (1 and 2). The value of \(A+B\) is \(1+2 = 3\). If \(A=2\): `22 x 22 = 484`. `ABA` would be `2B2`. This does not match.

Solution (B): 1. Look at the units: \(7 \times B\) must end in 2. 2. Test values for B: \(7 \times 1 = 7\), \(7 \times 2 = 14\), \(7 \times 3 = 21\), \(7 \times 4 = 28\), \(7 \times 5 = 35\), \(7 \times 6 = 42\). 3. The only digit that works is \(B = 6\). 4. The problem is now `A7 \times 6 = 3C2`. The result is in the 300s. 5. Test values for A: If \(A=4\): \(47 \times 6 = 282\). (Too small). If \(A=5\): \(57 \times 6 = 342\). This fits the pattern `3C2`, with \(C=4\). If \(A=6\): \(67 \times 6 = 402\). (Too large). 6. The only solution is \(A=5, B=6, C=4\). They are distinct. The digit A is 5.

Solution (D): 1. Divisible by 18 means divisible by 2 AND 9. 2. Divisible by 2: The last digit, X, must be even. X can be {0, 2, 4, 6, 8}. 3. Divisible by 9: The sum of digits `3+5+X+7+X = 15+2X` must be a multiple of 9. 4. Test the even X values: – \(X=0\): \(15 + 2(0) = 15\). (Not a multiple of 9). – \(X=2\): \(15 + 2(2) = 19\). (Not a multiple of 9). – \(X=4\): \(15 + 2(4) = 23\). (Not a multiple of 9). – \(X=6\): \(15 + 2(6) = 15 + 12 = 27\). (This is a multiple of 9). – \(X=8\): \(15 + 2(8) = 15 + 16 = 31\). (Not a multiple of 9). The only solution is \(X=6\).

Solution (B): 1. Units: \(Z+X = 8\) (no carry) or \(Z+X = 18\) (carry 1). 2. Tens: \(Y+Y + (\text{carry}) = 4\) or \(Y+Y + (\text{carry}) = 14\). 3. Hundreds: \(X+Z + (\text{carry}) = 8\). 4. From Hundreds: The carry from the tens must be 0, because if it were 1, \(X+Z=7\), which contradicts Units (8 or 18). 5. So, Carry from tens = 0. This means `X+Z = 8`. 6. This also means (from Units) that \(Z+X = 8\), so there is no carry from Units. 7. From Tens: \(Y+Y + 0 = 4\). \(2Y=4 \implies Y=2\). 8. We have \(Y=2\) and \(X+Z=8\). X and Z must be distinct and not 2. (e.g., X=1, Z=7, Y=2. \(127 + 721 = 848\). This works). 9. The sum is \(X+Y+Z = (X+Z) + Y = 8 + 2 = 10\).

Solution (C): This is from image 6. `X3Y + 5YX = 133X` 1. Hundreds: \(X+5 + (\text{carry}) = 13\). – If carry=0, \(X=8\). – If carry=1, \(X=7\). 2. Units: \(Y+X = X\) (no carry) or \(Y+X = 10+X\) (carry 1). Since Y is non-zero, \(Y=0\) is impossible. So \(Y+X = 10+X \implies Y=10\). Also impossible. *Correction:* `Y+X` ends in `X`. This means `Y=0` or `Y=10`. Since Y must be a non-zero digit, my assumption is wrong. Let’s re-read the image… `+ 5YX`. The image has `+ 5YX`. `X3Y + 5YX = 133X` Units: `Y+X = X`? No. `Y+X` must end in `X`. This means `Y=0` or `Y=10`. `Y` must be 0. But the image says “two different digits”. Let’s assume non-zero. Wait, the image for Q6 says `X3Y + 5YX = 1,33X`. This implies X is the *thousands* digit. Let’s re-read *that* image. `X3Y + 5YX = 133X`. Ah, `1,33X`. So `X=1`. Let’s check. `13Y + 5Y1 = 1331`. Units: `Y+1 = 1 \implies Y=0`. Tens: `3+Y = 3 \implies 3+0 = 3`. (Correct). Hundreds: `1+5 = 13`. (This is 6, not 13). The image Q6 is `X3Y + 5YX = 1,33X`. This must be `X=1`. `13Y + 5Y1 = 1331`. `Y+1 = 1 \implies Y=0`. `3+Y = 3 \implies 3+0 = 3`. `1+5 = 13`? No, `1+5=6`. The image `1,33X` must mean `C=1, D=3, E=3, F=X`. So the sum is `133X`. This is `1000 + 300 + 30 + X`. ` (100X + 30 + Y) + (500 + 10Y + X) = 1300 + 30 + X ` ` 101X + 11Y + 530 = 1330 + X ` ` 100X + 11Y = 800 ` We need an integer solution. If \(X=7\), `100(7) + 11Y = 800 \implies 700 + 11Y = 800 \implies 11Y = 100`. (No). If \(X=8\), `100(8) + 11Y = 800 \implies 800 + 11Y = 800 \implies 11Y = 0 \implies Y=0`. So `X=8, Y=0`. Check: `X3Y + 5YX = 830 + 508 = 1338`. Result `133X = 1338`. This works! `X=8`.

Solution (D): 1. Write the problem algebraically: \((100X + 30 + Y) + (500 + 10Y + X) = 1300 + 30 + X\). 2. Combine terms: \(101X + 11Y + 530 = 1330 + X\). 3. Subtract X and 530 from both sides: \(100X + 11Y = 800\). 4. Since \(11Y\) must be a multiple of 100 (because \(100X\) and \(800\) are), \(Y\) must be 0. 5. \(100X + 11(0) = 800 \implies 100X = 800 \implies X = 8\). 6. The digits are \(X=8\) and \(Y=0\). They are distinct. 7. Check: \(830 + 508 = 1338\). This matches the pattern `133X`.