Daily Math Puzzle: 2026-09-27

9 — Let's set up the cryptarithm vertically: P I E S + C A K E ---------- F O O D Each letter represents a unique digit from 0-9. P, C, F cannot be 0 as they are leading digits. We can break this down into four column additions, from right to left, noting carries (C1, C2, C3): 1. Units column: S + E = D + 10 * C1 (where C1 is the carry to the tens column, C1 can be 0 or 1) 2. Tens column: E + K + C1 = O + 10 * C2 (where C2 is the carry to the hundreds column, C2 can be 0 or 1) 3. Hundreds column: I + A + C2 = O + 10 * C3 (where C3 is the carry to the thousands column, C3 can be 0 or 1) 4. Thousands column: P + C + C3 = F Now, let's deduce the carries: * From column 4: P + C + C3 = F. Since PIES and CAKE are both 4-digit numbers, and FOOD is also a 4-digit number, there cannot be a carry out from the thousands column. This implies P + C + C3 must be less than 10. Since P, C, and F are single digits (and cannot be 0), C3 must be 0. If C3 were 1, P+C+1 would be at least 1+2+1=4, and at most 9+8+1=18. If it's 10 or more, FOOD would be a 5-digit number, which is not the case. So, C3 = 0. * From column 3: I + A + C2 = O + 10 * C3. Since C3 = 0, this simplifies to I + A + C2 = O. Since O is a single digit (and not 0), I + A + C2 must be less than 10. This means C2 must be 0. If C2 were 1, I+A+1 would be at least 1+2+1=4, which is possible for O. However, if C2=1, then I+A+1 would be O, but if O is a single digit, it must be <10. If O=I+A+C2 and O is the same letter in both column 2 and 3, then it must be the case that I+A+C2 < 10. This implies C2 = 0. So, we have crucial deductions: C3 = 0 and C2 = 0. Our simplified equations become: 1. S + E = D + 10 * C1 (C1 can be 0 or 1) 2. E + K + C1 = O 3. I + A = O 4. P + C = F (where P, C, F are distinct, and P, C, F are not 0, and P+C < 10) From (2) and (3), we have O = E + K + C1 and O = I + A. This implies I + A = E + K + C1. Since I, A, E, K, O are unique and none can be 0 (as they sum to O, or are part of O's sum), they must be distinct non-zero digits. Let's consider C1: * If C1 = 0: Then I + A = E + K. To satisfy this with unique, non-zero I,A,E,K, we need to find two disjoint pairs that sum to the same number (O). For example, if O=5, {1,4} and {2,3} sum to 5. So, (I,A) could be {1,4} and (E,K) could be {2,3}. Digits used: 1,2,3,4,5. Remaining: 0,6,7,8,9. For (P,C,F) from P+C=F, P,C from {6,7,8,9} would sum to at least 6+7=13, which cannot be a single digit F. This means C1=0 is not possible, as we cannot find valid P,C,F from the remaining digits. Therefore, C1 MUST be 1. Our final set of equations: 1. S + E = D + 10 (This means S+E must be 10 or more) 2. E + K + 1 = O 3. I + A = O 4. P + C = F (P, C, F are distinct, non-zero, and P+C < 10) From (2) and (3), I + A = E + K + 1. Also, O = I + A. Since I, A, E, K are distinct and non-zero, let's find a valid O. Minimum value for E+K+1 is 1+2+1=4, so O>=4. Similarly, I+A must be at least 1+2=3. Let's test possible values for O, looking for disjoint sets {I,A} and {E,K} satisfying I+A = E+K+1: * If O = 7: We need I+A=7 and E+K+1=7 (so E+K=6). * Possible pairs for {I,A} (sum 7): {1,6}, {2,5}, {3,4}. * Possible pairs for {E,K} (sum 6): {1,5}, {2,4}. * Let's try {I,A}={3,4} and {E,K}={1,5}. These sets are disjoint and satisfy the condition. So, we have O=7, {I,A}={3,4}, {E,K}={1,5}. (Digits used: 1,3,4,5,7) Now, let's find P, C, F using P + C = F. The remaining digits are {0,2,6,8,9}. P and C cannot be 0. * From the remaining digits, P=2, C=6 gives F=8 (2+6=8). This is a valid set {P,C,F} = {2,6,8}. (Digits used: 1,2,3,4,5,6,7,8) The final two remaining digits must be S and D: {0,9}. Finally, use S + E = D + 10: * E can be 1 or 5 (from {E,K}={1,5}). * If E=1: S + 1 = D + 10. With S,D as {0,9}: * If S=9: 9 + 1 = D + 10 => 10 = D + 10 => D = 0. This works! S=9, D=0. E=1. (All unique digits 0-9 are now assigned). Let's verify the complete assignment: * O = 7 * I = 3 (or 4), A = 4 (or 3) * E = 1, K = 5 (or 1) * P = 2 (or 6), C = 6 (or 2) * F = 8 * S = 9 * D = 0 Assigned unique digits: 0=D, 1=E, 2=P, 3=I, 4=A, 5=K, 6=C, 7=O, 8=F, 9=S. Let's check the addition: P I E S -> 2 3 1 9 + C A K E -> 6 4 5 1 ---------- F O O D -> 8 7 7 0 1. Units: S + E = 9 + 1 = 10. D=0, C1=1. (Correct) 2. Tens: E + K + C1 = 1 + 5 + 1 = 7. O=7, C2=0. (Correct) 3. Hundreds: I + A + C2 = 3 + 4 + 0 = 7. O=7, C3=0. (Correct) 4. Thousands: P + C + C3 = 2 + 6 + 0 = 8. F=8. (Correct) The solution is consistent, and 'S' represents the digit 9. The final answer is **9**.