Puzzle for December 22, 2019 ( )
Scratchpad
Cind 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
Subtract the left and right sides of eq.4 from the left and right sides of eq.3, respectively: A + C – (A – B) = D + F – (D – F) which is equivalent to A + C – A + B = D + F – D + F which becomes C + B = 2×F which may be written as eq.3a) B + C = 2×F
Hint #2
Add C to each side of eq.2: B + F + C = A – C + E + C which may be written as B + C + F = A + E In the equation above, replace B + C with 2×F (from eq.3a): 2×F + F = A + E which becomes eq.2a) 3×F = A + E
Hint #3
eq.1 may be re-written as: B + C + A + E + D + F = 36 In the above equation, replace B + C with 2×F (from eq.3a), and replace A + E with 3×F (from eq.2a): 2×F + 3×F + D + F = 36 which becomes 6×F + D = 36 Subtract 6×F from both sides: 6×F + D – 6×F = 36 – 6×F which becomes eq.1a) D = 36 – 6×F
Hint #4
To make eq.1a true, check several possible values for F and D: If F = 6 then D = 36 – 6×6 = 36 – 36 = 0 If F = 7 then D = 36 – 6×7 = 36 – 42 = –6 If F > 7 then D < –6 If F = 5 then D = 36 – 6×5 = 36 – 30 = 6 If F = 4 then D = 36 – 6×4 = 36 – 24 = 12 If F < 4 then D > 12 Since D and F must be one-digit non-negative integers, then either: F = 6 and D = 0 or: F = 5 and D = 6
Hint #5
Begin checking: F = 6, and D = 0 ... Substituting 0 for D, and 6 for F in eq.3 would yield: A + C = 0 + 6 which would become eq.1b) A + C = 6
Hint #6
Continue checking: F = 6, and D = 0 ... Re-writing eq.1 would yield: B + E + A + C + D + F = 36 Substituting 6 for A + C, 0 for D, and 6 for F would yield: B + E + 6 + 0 + 6 = 36 which would become B + E + 12 = 36 Subtracting 12 from each side would yield: B + E + 12 – 12 = 36 – 12 which would make eq.1c) B + E = 24
Hint #7
Finish checking: F = 6, and D = 0 ... Since B and E are one-digit non-negative integers, then: B + E ≤ 18 which means B + E ≠ 24 (from eq.1c) and also means F ≠ 6 and D ≠ 0 and therefore makes F = 5 and D = 6
Hint #8
Substitute 5 for F in eq.3a: B + C = 2×5 which becomes B + C = 10 Subtract C from both sides of the equation above: B + C – C = 10 – C which becomes eq.3b) B = 10 – C
Hint #9
Substitute 10 – C for B (from eq.3b), and 6 for D in eq.5: C = 10 – C – 6 which becomes C = 4 – C Add C to each side of the above equation: C + C = 4 – C + C which means 2×C = 4 Divide both sides by 2: 2×C ÷ 2 = 4 ÷ 2 which makes C = 2
Hint #10
Substitute 2 for C in eq.3b: B = 10 – 2 which makes B = 8
Hint #11
Substitute 8 for B, 6 for D, and 5 for F in eq.4: A – 8 = 6 – 5 which becomes A – 8 = 1 Add 8 to each side of the above equation: A – 8 + 8 = 1 + 8 which makes A = 9
Solution
Substitute 8 for B, 5 for F, 9 for A, and 2 for C in eq.2: 8 + 5 = 9 – 2 + E which becomes 13 = 7 + E Subtract 7 from each side of the above equation: 13 – 7 = 7 + E – 7 which makes 6 = E and ABCDEF = 982665