Puzzle for March 6, 2021 ( )
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.
* CD and EF are 2-digit numbers (not C×D or E×F).
Scratchpad
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Hint #1
Add F to both sides of eq.5: C – F + F = E – (C – A) + F which becomes C = E – C + A + F which may be written as eq.5a) C = A + E + F – C In eq.5a, replace A + E + F with B + C (from eq.3): C = B + C – C which makes C = B
Hint #2
In eq.2, replace B with C: C = C + D Subtract C from each side of the above equation: C – C = C + D – C which means 0 = D
Hint #3
Add A to both sides of eq.4: A – C + A = E – A + A which becomes eq.4a) 2×A – C = E In eq.5a, substitute 2×A – C for E (from eq.4a): C = A + 2×A – C + F – C which becomes C = 3×A – 2×C + F In the above equation, subtract 3×A from both sides, and add 2×C to both sides: C – 3×A + 2×C = 3×A – 2×C + F – 3×A + 2×C which becomes eq.5b) 3×C – 3×A = F
Hint #4
eq.6 may be written as: E + 10×E + F = 10×C + D – E – F which becomes 11×E + F = 10×C + D – E – F Add E and F to both sides of the above equation: 11×E + F + E + F = 10×C + D – E – F + E + F which becomes 12×E + 2×F = 10×C + D which my be written as eq.6a) 12×E + 2×F = 5×(2×C) + D
Hint #5
Add C to both sides of eq.5a: C + C = A + E + F – C + C which becomes 2×C = A + E + F In eq.6a, substitute A + E + F for 2×C, and 0 for D: 12×E + 2×F = 5×(A + E + F) + 0 which becomes 12×E + 2×F = 5×A + 5×E + 5×F Subtract 5×E and 2×F from both sides of the above equation: 12×E + 2×F – 5×E – 2×F = 5×A + 5×E + 5×F – 5×E – 2×F which becomes eq.6b) 7×E = 5×A + 3×F
Hint #6
Substitute (2×A – C) for E (from eq.4a), and (3×C – 3×A) for F (from eq.5b) in eq.6b: 7×(2×A – C) = 5×A + 3×(3×C – 3×A) which becomes 14×A – 7×C = 5×A + 9×C – 9×A which becomes 14×A – 7×C = 9×C – 4×A Add 4×A and 7×C to both sides of the above equation: 14×A – 7×C + 4×A + 7×C = 9×C – 4×A + 4×A + 7×C which makes 18×A = 16×C Divide both sides by 16: 18×A ÷ 16 = 16×C ÷ 16 which makes 1⅛×A = C and also makes B = C = 1⅛×A
Hint #7
Substitute 1⅛×A for C in eq.4a: 2×A – 1⅛×A = E which makes ⅞×A = E
Hint #8
Substitute (1⅛×A) for C in eq.5b: 3×(1⅛×A) – 3×A = F which becomes 3⅜×A – 3×A = F which makes ⅜×A = F
Solution
Substitute 1⅛×A for B and C, 0 for D, ⅞×A for E, and ⅜×A for F in eq.1: A + 1⅛×A + 1⅛×A + 0 + ⅞×A + ⅜×A = 36 which simplifies to 4½×A = 36 Divide both sides of the above equation by 4½: 4½×A ÷ 4½ = 36 ÷ 4½ which means A = 8 making B = C = 1⅛×A = 1⅛ × 8 = 9 E = ⅞×A = ⅞ × 8 = 7 F = ⅜×A = ⅜ × 8 = 3 and ABCDEF = 899073