Puzzle for May 1, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
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Hint #1
Add D to both sides of eq.3: A + D + D = C – D + E + D which becomes eq.3a) A + 2×D = C + E In eq.4, add C to both sides, and subtract D from both sides: D + E + C – D = A – C + F + C – D which becomes E + C = A + F – D which may be written as eq.4a) C + E = A – D + F
Hint #2
In eq.3a, replace C + E with A – D + F (from eq.4a): A + 2×D = A – D + F In the equation above, subtract A from both sides, and add D to both sides: A + 2×D – A + D = A – D + F – A + D which simplifies to 3×D = F
Hint #3
In eq.4a, replace C + E with B – C + F (from eq.2): B – C + F = A – D + F In the above equation, add C and D to both sides, and subtract F from both sides: B – C + F + C + D – F = A – D + F + C + D – F which becomes eq.4b) B + D = A + C
Hint #4
Add E, D, and F to both sides of eq.5: A – E + E + D + F = B + C – D – F + E + D + F which becomes eq.5a) A + D + F = B + C + E In eq.5a, substitute B – C + F for C + E (from eq.2): A + D + F = B + B – C + F which becomes A + D + F = 2×B – C + F In the above equation, subtract F from both sides, and add C to both sides: A + D + F – F + C = 2×B – C + F – F + C which becomes A + D + C = 2×B which may be written as eq.5b) A + C + D = 2×B
Hint #5
Substitute B + D for A + C (from eq.4b) into eq.5b: B + D + D = 2×B which becomes B + 2×D = 2×B Subtract B from each side of the equation above: B + 2×D – B = 2×B – B which makes 2×D = B
Hint #6
Substitute B + D for A + C (from eq.4b) in eq.6: eq.6a) A × B = B + D + F
Hint #7
Substitute 2×D for B, and 3×D for F in eq.6a: A × 2×D = 2×D + D + 3×D which becomes A × 2×D = 6×D Divide both sides of the above equation by 2×D (assumes D ≠ 0): A × 2×D ÷ 2×D = 6×D ÷ 2×D which makes A = 3
Hint #8
Begin confirming: D ≠ 0 ... If D = 0, then: B = 2×D = 2×0 = 0 F = 3×D = 3×0 = 0 Substituting 0 for B and F in eq.2 would yield: C + E = 0 – C + 0 which would make C + E = –C The only value for C and E that would make the above equation true is: C = 0 and E = 0
Hint #9
Finish confirming: D ≠ 0 ... Substituting 0 for B, C, D, E, and F in eq.1 would yield: A + 0 + 0 + 0 + 0 + 0 = 14 which would make A = 14 Since A is a one-digit integer, then: A ≠ 14 and therefore means: D ≠ 0
Hint #10
eq.1 may be written as: A + D + F + B + C + E = 14 Substitute A + D + F for B + C + E (from eq.5a) into the above equation: A + D + F + A + D + F = 14 which becomes 2×(A + D + F) = 14 Divide both sides of the above equation by 2: 2×(A + D + F) ÷ 2 = 14 ÷ 2 which becomes eq.1a) A + D + F = 7
Hint #11
Substitute 3 for A, and 3×D for F in eq.1a: 3 + D + 3×D = 7 which becomes 3 + 4×D = 7 Subtract 3 from each side of the equation above: 3 + 4×D – 3 = 7 – 3 which makes 4×D = 4 Divide both sides by 4: 4×D ÷ 4 = 4 ÷ 4 which makes D = 1 making B = 2×D = 2×1 = 2 F = 3×D = 3×1 = 3
Hint #12
Substitute 2 for B, 1 for D, and 3 for A in eq.4b: 2 + 1 = 3 + C which makes 3 = 3 + C Subtract 3 from each side of the above equation: 3 – 3 = 3 + C – 3 which makes 0 = C
Solution
Substitute 3 for A, 1 for D, and 0 for C in eq.3a: 3 + 2×1 = 0 + E which becomes 3 + 2 = E which makes 5 = E and makes ABCDEF = 320153