Daily Math Puzzle: 2026-10-01

12 — The puzzle translates to the cryptarithmetic sum: B U Y + B U Y + 1 ------- P A I D 1. From the leftmost column (thousands place), since B is a single digit and the sum results in a four-digit number `PAID`, there must be a carry-over. The maximum sum of B+B+carry_from_U is 9+9+1=19. Therefore, P must be 1. 2. Let C1, C2, C3 be the carries from right to left (units, tens, hundreds place). - Y + Y + 1 = D (+ 10 * C1) - U + U + C1 = I (+ 10 * C2) - B + B + C2 = A (+ 10 * C3) - C3 = P Since P=1, we know C3=1. This implies B+B+C2 must be 10 or more. Because B is the first digit, B cannot be 0 or 1 (as P=1). Let's systematically search for solutions that minimize P+A+I+D. P is fixed at 1. Case 1: B=5. 5 U Y + 5 U Y + 1 --------- 1 A I D From B+B+C2 = A+10: 5+5+C2 = A+10 => 10+C2 = A+10 => C2 = A. Since A and P (1) must be unique digits, A cannot be 1. Thus, C2 cannot be 1. So, C2 must be 0. This means A=0. So far: P=1, B=5, A=0. Used digits: {0,1,5}. From C2=0, we know U+U+C1 < 10. Now consider C1 (carry from Y+Y+1): Subcase 1.1: C1=0. This means Y+Y+1 < 10, so Y can be at most 4 (4+4+1=9). Also, U+U+C1=I => U+U=I. Since C2=0, U+U < 10. So U can be at most 4. Letters available: {2,3,4,6,7,8,9} (excluding 0,1,5). If U=2, then I=4. Used digits: {0,1,2,4,5}. Remaining for Y,D: {3,6,7,8,9}. From Y+Y+1 = D (C1=0 implies D<10): If Y=3, then 3+3+1 = 7. So D=7. This gives the assignment: B=5, U=2, Y=3 (BUY = 523) and P=1, A=0, I=4, D=7 (PAID = 1047). All digits {5,2,3,1,0,4,7} are unique. Let's verify: 523 + 523 + 1 = 1047. This is a valid solution. Sum of P+A+I+D = 1+0+4+7 = 12. Let's check for other solutions to see if a smaller sum of P+A+I+D is possible: If we continue Subcase 1.1 (B=5, A=0, C2=0, C1=0): We found U=2, I=4, and Y=3, D=7, which gives Sum(P+A+I+D) = 12. If Y is larger than 3, Y+Y+1 will be 10 or more (e.g., if Y=6, 6+6+1=13, so C1 would be 1, contradicting C1=0). Subcase 1.2: C1=1. This means Y+Y+1 >= 10. So Y must be at least 5. Also, U+U+C1=I => U+U+1=I. Since C2=0, U+U+1 < 10. So U can be at most 4. U cannot be 5 (B=5). If U=3, then 3+3+1 = 7. So I=7. Used digits: {0,1,3,5,7}. Remaining for Y,D: {2,4,6,8,9}. From Y+Y+1 = D+10 (C1=1 implies D<10): Y must be at least 5. Y cannot be 5 (B=5) or 7 (I=7). If Y=6, then 6+6+1=13. So D=3. But U=3, so D cannot be 3. (Digits must be unique). If Y=8, then 8+8+1=17. So D=7. But I=7, so D cannot be 7. If Y=9, then 9+9+1=19. So D=9. But Y=9, so D cannot be 9. No solution in this subcase. Continuing with higher values for B: Case 2: B=6. P=1, B=6. A=C2+2. (from B+B+C2=A+10). If C2=0, A=2. Used: {0,1,2,6}. Then U+U+C1 < 10. (C1=0, U+U=I or C1=1, U+U+1=I, where I<10). A valid assignment is: B=6, U=4, Y=3 (BUY=643) => P=1, A=2, I=8, D=7 (PAID=1287). Sum(P+A+I+D) = 1+2+8+7 = 18. Case 3: B=6. P=1, B=6. If C2=1, A=3. Used: {1,3,6}. Then U+U+C1 >= 10. (C1=0, U+U=I+10 or C1=1, U+U+1=I+10, where I<10). A valid assignment is: B=6, U=5, Y=4 (BUY=654) => P=1, A=3, I=0, D=9 (PAID=1309). Sum(P+A+I+D) = 1+3+0+9 = 13. Case 4: B=7. P=1, B=7. A=C2+4. If C2=0, A=4. Used: {1,4,7}. A valid assignment is: B=7, U=3, Y=2 (BUY=732) => P=1, A=4, I=6, D=5 (PAID=1465). Sum(P+A+I+D) = 1+4+6+5 = 16. Comparing the sums of P+A+I+D found: - From BUY=523 (PAID=1047): 12 - From BUY=643 (PAID=1287): 18 - From BUY=654 (PAID=1309): 13 - From BUY=732 (PAID=1465): 16 The smallest possible sum of the digits P+A+I+D found is 12. The correct choice is 12.