Puzzle for December 12, 2020 ( )
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
eq.4 may be written as: B + E + F – C – D = A – B + C + D – E In the above equation, replace B + E + F with A – B + C (from eq.5): A – B + C – C – D = A – B + C + D – E which becomes A – B – D = A – B + C + D – E Add B, D, and E to both sides: A – B – D + B + D + E = A – B + C + D – E + B + D + E which becomes A + E = A + C + 2×D Subtract A from both sides: A + E – A = A + C + 2×D – A which becomes eq.4a) E = C + 2×D
Hint #2
In eq.2, replace E with C + 2×D (from eq.4a): C + 2×D = C + D Subtract C and D from each side of the equation above: C + 2×D – C – D = C + D – C – D which makes D = 0
Hint #3
In eq.2, substitute 0 for D: E = C + 0 which makes E = C
Hint #4
Substitute 0 for D in eq.3: 0 + E = B + F which makes E = B + F and also makes eq.3a) C = E = B + F
Hint #5
Substitute B + F for C and E (from eq.3a) in eq.5: A – B + B + F = B + B + F + F which becomes A + F = 2×B + 2×F Subtract F from each side of the above equation: A + F – F = 2×B + 2×F – F which becomes eq.5a) A = 2×B + F
Hint #6
Substitute (B + F) for C and E (from eq.3a), and (2×B + F) for A (from eq.5a) in eq.6: (B + F) × (B + F) = ((2×B + F) × F) + B which becomes B×B + B×F + F×B + F×F = 2×B×F + F×F + B which is equivalent to B² + 2×B×F + F² = 2×B×F + F² + B Subtract 2×B×F and F² from both sides of the equation above: B² + 2×B×F + F² – 2×B×F – F² = 2×B×F + F² + B – 2×B×F – F² which simplifies to B² = B making B = 1 or B = 0
Hint #7
Begin checking: B = 0 ... Substituting 0 for B in eq.3a would yield: C = E = 0 + F which would make C = E = F Substituting 0 for B in eq.5a would yield: A = 2×0 + F which would make A = F
Hint #8
Finish checking: B = 0 ... Substituting F for A and C and E, and 0 for B and D in eq.1 would yield: F + 0 + F + 0 + F + F = 33 which would become 4×F = 33 Dividing both sides of the equation above by 4 would yield: 4×F ÷ 4 = 33 ÷ 4 which would make F = 8¼ Since F is an integer, then F ≠ 8¼ which means B ≠ 0 and therefore makes B = 1
Hint #9
Substitute 1 for B in eq.3a: eq.3b) C = E = 1 + F
Hint #10
Substitute 1 for B in eq.5a: A = 2×1 + F which makes eq.5b) A = 2 + F
Solution
Substitute 2 + F for A (from eq.5b), 1 for B, and 1 + F for C and E (from eq.3b), and 0 for D in eq.1: 2 + F + 1 + 1 + F + 0 + 1 + F + F = 33 which becomes 5 + 4×F = 33 Subtract 5 from both sides of the equation above: 5 + 4×F – 5 = 33 – 5 which makes 4×F = 28 Divide both sides by 4: 4×F ÷ 4 = 28 ÷ 4 which means F = 7 making A = 2 + F = 2 + 7 = 9 (from eq.5b) C = E = 1 + F = 1 + 7 = 8 (from eq.3b) and ABCDEF = 918087