Puzzle for February 19, 2022 ( )
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.
* BC, DE, and EF are 2-digit numbers (not B×C, D×E, or E×F).
Scratchpad
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Hint #1
Add D and C to both sides of eq.5: B + E – D + D + C = D + F – C + D + C which becomes B + E + C = 2×D + F which may be written as B + C + E = 2×D + F In the above equation, replace B + C with A + E (from eq.3): A + E + E = 2×D + F which becomes eq.5a) A + 2×E = 2×D + F
Hint #2
Add F to both sides of eq.4: E + F + F = A + C – F + F which becomes eq.4a) E + 2×F = A + C In eq.4a, replace A + C with D (from eq.2): eq.4b) E + 2×F = D
Hint #3
In eq.5a, substitute (E + 2×F) for D (from eq.4b): A + 2×E = 2×(E + 2×F) + F which becomes A + 2×E = 2×E + 4×F + F which becomes A + 2×E = 2×E + 5×F Subtract 2×E frfom each side of the equation above: A + 2×E – 2×E = 2×E + 5×F – 2×E which makes A = 5×F
Hint #4
Substitute 5×F for A in eq.4a: E + 2×F = 5×F + C Subtract 5×F from each side of the above equation: E + 2×F – 5×F = 5×F + C – 5×F which becomes eq.4c) E – 3×F = C
Hint #5
Substitute E – 3×F for C (from eq.4c), and 5×F for A in eq.3: B + E – 3×F = 5×F + E In the equation above, subtract E from both sides, and add 3×F to both sides: B + E – 3×F – E + 3×F = 5×F + E – E + 3×F which simplifies to B = 8×F
Hint #6
eq.6 may be written as: A + 10×B + C – (10×E + F) = 10×D + E + F which becomes A + 10×B + C – 10×E – F = 10×D + E + F Add 10×E and F to both sides of the above equation: A + 10×B + C – 10×E – F + 10×E + F = 10×D + E + F + 10×E + F which becomes eq.6a) A + 10×B + C = 10×D + 11×E + 2×F
Hint #7
Substitute 5×F for A, (8×F) for B, E – 3×F for C (from eq.4c), and (E + 2×F) for D (from eq.4b) in eq.6a: 5×F + 10×(8×F) + E – 3×F = 10×(E + 2×F) + 11×E + 2×F which becomes 5×F + 80×F + E – 3×F = 10×E + 20×F + 11×E + 2×F which becomes 82×F + E = 21×E + 22×F Subtract E and 22×F from each side of the above equation: 82×F + E – E – 22×F = 21×E + 22×F – E – 22×F which makes 60×F = 20×E Divide both sides by 20: 60×F ÷ 20 = 20×E ÷ 20 which makes 3×F = E
Hint #8
Substitute 3×F for E in eq.4b: 3×F + 2×F = D which makes 5×F = D
Hint #9
Substitute 3×F for E in eq.4c: 3×F – 3×F = C which makes 0 = C
Solution
Substitute 5×F for A and D, 8×F for B, 0 for C, and 3×F for E in eq.1: 5×F + 8×F + 0 + 5×F + 3×F + F = 22 which simplifies to 22×F = 22 Divide both sides of the above equation by 22: 22×F ÷ 22 = 22 ÷ 22 which means F = 1 making A = D = 5×F = 5 × 1 = 5 B = 8×F = 8 × 1 = 8 E = 3×F = 3 × 1 = 3 and ABCDEF = 580531