Puzzle for December 27, 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
Subtract the left and right sides of eq.3 from the left and right sides of eq.2, respectively: B + E – (E + F – B) = A + C + D – (B + C + D) which is equivalent to B + E – E – F + B = A + C + D – B – C – D which becomes 2×B – F = A – B Add B to both sides of the equation above: 2×B – F + B = A – B + B which becomes eq.3a) 3×B – F = A
Hint #2
eq.5 may be written as: B = (A + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (A + C + F) ÷ 3 which becomes eq.5a) 3×B = A + C + F
Hint #3
In eq.3a, replace 3×B with A + C + F (from eq.5a): A + C + F – F = A which becomes A + C = A Subtract A from each side of the equation above: A + C – A = A – A which means C = 0
Hint #4
In eq.4, substitute 3×B – F for A (from eq.3a): 3×B – F + E = B + F In the above equation, subtract 3×B from each side, and add F to each side: 3×B – F + E – 3×B + F = B + F – 3×B + F which becomes E = 2×F – 2×B which may be written as eq.4a) E = 2×(F – B)
Hint #5
Substitute E ÷ A for F – B (from eq.6) in eq.4a: E = 2×(E ÷ A) which may be equivalently re-written as eq.4b) E = (2 ÷ A) × E To make eq.4b true, then either E = 0 or A = 2
Hint #6
Begin checking: E = 0 ... Substituting 0 for E in eq.4a would yield: 0 = 2×(F – B) which would make 0 = F – B which would make B = F
Hint #7
Continue checking: E = 0 ... Substituting 0 for E, and F for B in eq.3 would yield: 0 + F – F = B + C + D which would make 0 = B + C + D Since B, C, and D are non-negative, the above equation would make B = C = D = 0 and would also make F = B = 0
Hint #8
Finish checking: E = 0 ... Substituting 0 for E, B, C, D, and F in eq.1 would yield: A + 0 + 0 + 0 + 0 + 0 = 29 which would make A = 29 Since A must be a one-digit integer, then A ≠ 29 which means E ≠ 0 and therefore makes A = 2
Hint #9
Substitute 2 for A in eq.3a: 3×B – F = 2 In the equation above, add F to both sides, and subtract 2 from both sides: 3×B – F + F – 2 = 2 + F – 2 which becomes eq.3b) 3×B – 2 = F
Hint #10
Substitute 3×B – 2 for F (from eq.3b) in eq.4a: E = 2×(3×B – 2 – B) which becomes E = 2×(2×B – 2) which becomes eq.4b) E = 4×B – 4
Hint #11
Substitute 4×B – 4 for E (from eq.4b), 2 for A, and 0 for C in eq.2: B + 4×B – 4 = 2 + 0 + D which becomes 5×B – 4 = 2 + D Subtract 2 from each side of the equation above: 5×B – 4 – 2 = 2 + D – 2 which becomes eq.2a) 5×B – 6 = D
Solution
Substitute 2 for A, 0 for C, 5×B – 6 for D (from eq.2a), 4×B – 4 for E (from eq.4b), and 3×B – 2 for F (from eq.3b) in eq.1: 2 + B + 0 + 5×B – 6 + 4×B – 4 + 3×B – 2 = 29 which simplifies to 13×B – 10 = 29 Add 10 to both sides of the equation above: 13×B – 10 + 10 = 29 + 10 which makes 13×B = 39 Divide both sides of the above equation by 13: 13×B ÷ 13 = 39 ÷ 13 which means B = 3 making D = 5×B – 6 = 5×3 – 6 = 15 – 6 = 9 (from eq.2a) E = 4×B – 4 = 4×3 – 4 = 12 – 4 = 8 (from eq.4b) F = 3×B – 2 = 3×3 – 2 = 9 – 2 = 7 (from eq.3b) and ABCDEF = 230987