Puzzle for November 24, 2019 ( )
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 C and D to both sides of eq.5: B – C + C + D = A – D + C + D which becomes B + D = A + C In eq.2, replace A + C with B + D: B + D + F = B + D Subtract B and D from both sides of the above equation: B + D – B – D = B + D + F – B – D which simplifies to 0 = F
Hint #2
Subtract C from each side of eq.3: A – B + C – C = B – C which becomes A – B = B – C In the equation above, replace B – C with A – D (from eq.5): A – B = A – D Subtract A from both sides of the above equation: A – B – A = A – D – A which makes –B = –D which means B = D
Hint #3
In eq.4, replace F with 0: C + 0 = A + E which becomes eq.4a) C = A + E
Hint #4
Add E to both sides of eq.6: (A × C) – E + E = A + (C × E) + E which becomes (A × C) = (C × E) + A + E In the above equation, substitute C for A + E (from eq.4a): eq.6a) (A × C) = (C × E) + C
Hint #5
Subtract (C × E) from both sides of eq.6a: (A × C) – (C × E) = (C × E) + C – (C × E) which becomes (A × C) – (C × E) = C which may be written as eq.6b) C × (A – E) = C To make eq.6b true, then: C = 0 and / or A – E = 1
Hint #6
Begin checking: C = 0 ... Substituting 0 for C in eq.4a would yield: 0 = A + E Since A and E must be non-negative integers, the above equation would make: A = 0 and E = 0
Hint #7
Finish checking: C = 0 ... Substituting B for D, and 0 for A, C, E, and F in eq.1 would make: 0 + B + 0 + B + 0 + 0 = 32 which would make 2×B = 32 Dividing both sides of the equation above by 2 would yield: 2×B ÷ 2 = 32 ÷ 2 which would make B = 16 Since B must be a one-digit integer, then: B ≠ 16 which means C ≠ 0 and therefore means eq.6c) A – E = 1
Hint #8
Add E to both sides of eq.6c: A – E + E = 1 + E which makes eq.6d) A = 1 + E Substitute 1 + E for A (from eq.6d) in eq.4a: C = 1 + E + E which makes eq.4b) C = 1 + 2×E
Hint #9
Substitute 1 + E for A (from eq.6d), and 1 + 2×E for C (from eq.4b) in eq.3: 1 + E – B + 1 + 2×E = B which becomes 2 + 3×E – B = B Add B to both sides of the equation above: 2 + 3×E – B + B = B + B which becomes 2 + 3×E = 2×B Divide both sides by 2: (2 + 3×E) ÷ 2 = 2×B ÷ 2 which makes 1 + 1½×E = B and also makes eq.3a) D = B = 1 + 1½×E
Solution
Substitute 1 + E for A (from eq.6d), 1 + 1½×E for B and D (from eq.3a), 1 + 2×E for C (from eq.4b), and 0 for F in eq.1: 1 + E + 1 + 1½×E + 1 + 2×E + 1 + 1½×E + E + 0 = 32 which simplifies to 4 + 7×E = 32 Subtract 4 from both sides of the above equation: 4 + 7×E – 4 = 32 – 4 which makes 7×E = 28 Divide both sides by 7: 7×E ÷ 7 = 28 ÷ 7 which means E = 4 making A = 1 + E = 1 + 4 = 5 (from eq.6d) B = D = 1 + 1½×E = 1 + (1½ × 4) = 1 + 6 = 7 (from eq.3a) C = 1 + 2×E = 1 + 2×4 = 1 + 8 = 9 (from eq.4b) and ABCDEF = 579740