Puzzle for October 23, 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, CD, and DE are 2-digit numbers (not B×C, C×D, or D×E).
Scratchpad
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Hint #1
Add C, A, and F to both sides of eq.2: F - C + C + A + F = D - A - F + C + A + F which becomes 2×F + A = D + C which may be written as eq.2a) A + 2×F = C + D Add F to both sides of eq.3: A + B - C - D + F + F = D + E - F + F which becomes A + B - C - D + 2×F = D + E which may be written as eq.3a) A + 2×F + B - C - D = D + E
Hint #2
In eq.3a, replace A + 2×F with C + D (from eq.2a): C + D + B - C - D = D + E which becomes eq.3b) B = D + E
Hint #3
eq.4 may be written as: 10×B + C - (10×C + D) = 10×D + E which becomes 10×B + C - 10×C - D = 10×D + E which becomes 10×B - 9×C - D = 10×D + E Add D to both sides of the above equation: 10×B - 9×C - D + D = 10×D + E + D which becomes eq.4a) 10×B - 9×C = 11×D + E
Hint #4
In eq.4a, substitute (D + E) for B (from eq.3b): 10×(D + E) - 9×C = 11×D + E which becomes 10×D + 10×E - 9×C = 11×D + E Subtract 10×D and E from each side of the above equation: 10×D + 10×E - 9×C - 10×D - E = 11×D + E - 10×D - E which becomes eq.4b) 9×E - 9×C = D
Hint #5
Divide both sides of eq.4b by 9: (9×E - 9×C) ÷ 9 = D ÷ 9 which becomes eq.4c) E - C = D ÷ 9 Since C, D, and E are one-digit non-negative integers, then in eq.4c: E - C = an integer which makes D ÷ 9 = an integer which means D = 0 or D = 9
Hint #6
Begin checking: D = 9 ... Substituting 9 for D in eq.4c would yield: E - C = 9 ÷ 9 which would make E - C = 1 Adding C to both sides of the above equation would yield: E - C + C = 1 + C which would make eq.4d) E = 1 + C
Hint #7
Finish checking: D = 9 ... Substituting 9 for D, and 1 + C for E (from eq.4d) into eq.3b would yield: B = 9 + 1 + C which would make B = 10 + C which would mean B ≥ 10 Since B is a one-digit non-negative integer, then: B < 10 which means D ≠ 9 and therefore makes D = 0
Hint #8
Substitute 0 for D in eq.3b: B = 0 + E which makes B = E
Hint #9
Substitute 0 for D in eq.4c: E - C = 0 ÷ 9 which becomes E - C = 0 Add C to both sides of the above equation: E - C + C = 0 + C which makes E = C and also makes E = C = B
Hint #10
Substitute 0 for D in eq.2: F - C = 0 - A - F Add C, A, and F to both sides of the equation above: F - C + C + A + F = 0 - A - F + C + A + F which simplifies to eq.2b) 2×F + A = C
Hint #11
eq.5 may be written as: A = (B + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + E + F) ÷ 3 which becomes eq.5a) 3×A = B + E + F
Hint #12
In eq.5a, substitute C for B and E: 3×A = C + C + F which becomes eq.5b) 3×A = 2×C + F
Hint #13
Substitute (2×F + A) for C (from eq.2b) into eq.5b: 3×A = 2(2×F + A) + F which becomes 3×A = 4×F + 2×A + F which becomes 3×A = 5×F + 2×A Subtract 2×A from each side of the equation above: 3×A - 2×A = 5×F + 2×A - 2×A which makes A = 5×F
Hint #14
Substitute 5×F for A in eq.2b: 2×F + 5×F = C which makes 7×F = C and also makes 7×F = C = B = E
Solution
Substitute 5×F for A, 7×F for B and C and E, and 0 for D in eq.1: 5×F + 7×F + 7×F + 0 + 7×F + F = 27 which simplifies to 27×F = 27 Divide both sides of the above equation by 27: 27×F ÷ 27 = 27 ÷ 27 which means F = 1 making A = 5×F = 5 × 1 = 5 B = C = E = 7×F = 7 × 1 = 7 and ABCDEF = 577071