Puzzle for February 24, 2019 ( )
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.
* AB, CD, DE, and EF are 2-digit numbers (not A×B, C×D, D×E, or E×F).
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Hint #1
Add E + F to both sides of eq.2: D - E + E + F = E - F + E + F which becomes eq.2a) D + F = 2×E
Hint #2
Add A + D to both sides of eq.5: E + F - A + A + D = A - D + A + D which becomes E + F + D = 2×A which may be written as E + D + F = 2×A In the above equation, replace D + F with 2×E (from eq.2a): E + 2×E = 2×A which becomes 3×E = 2×A Divide both sides by 2: 3×E ÷ 2 = 2×A ÷ 2 which makes eq.5a) 1½×E = A
Hint #3
Subtract F from both sides of eq.4: B + D - F = E + F - F which becomes B + D - F = E In eq.2a, replace E with (B + D - F): D + F = 2×(B + D - F) which is equivalent to D + F = 2×B + 2×D - 2×F Add (2×F - 2×D) to both sides of the above equation: D + F + (2×F - 2×D) = 2×B + 2×D - 2×F + (2×F - 2×D) which simplifies to eq.2b) 3×F - D = 2×B
Hint #4
Subtract C from both sides of eq.3: C + F - C = D - C which becomes eq.3a) F = D - C In eq.2b, substitute (D - C) for F: 3×(D - C) - D = 2×B which is the same as 3×D - 3×C - D = 2×B which becomes eq.2c) 2×D - 3×C = 2×B
Hint #5
eq.6 may be written as: 10×A + B + 10×C + D + E + F = 10×D + E + 10×E + F Subtract D, E, and F from each side of the above equation: 10×A + B + 10×C + D + E + F - D - E - F = 10×D + E + 10×E + F - D - E - F which becomes 10×A + B + 10×C = 9×D + 10×E Substitute (1½×E) for A (from eq.5a): 10×(1½×E) + B + 10×C = 9×D + 10×E which becomes 15×E + B + 10×C = 9×D + 10×E Subtract 10×E from each side: 15×E + B + 10×C - 10×E = 9×D + 10×E - 10×E which becomes eq.6a) 5×E + B + 10×C = 9×D
Hint #6
In eq.4, replace D with C + F (from eq.3): B + C + F = E + F Subtract F from both sides: B + C + F - F = E + F - F which becomes eq.4a) B + C = E Substitute (B + C) for E in eq.6a: 5×(B + C) + B + 10×C = 9×D which is the same as 5×B + 5×C + B + 10×C = 9×D which becomes eq.6b) 6×B + 15×C = 9×D
Hint #7
Divide both sides of eq.6b by 3: (6×B + 15×C) ÷ 3 = 9×D ÷ 3 which becomes 2×B + 5×C = 3×D In the above equation, substitute 2×D - 3×C for 2×B (from eq.2c): 2×D - 3×C + 5×C = 3×D which becomes 2×D + 2×C = 3×D Subtract 2×D from both sides: 2×D + 2×C - 2×D = 3×D - 2×D which makes eq.2d) 2×C = D
Hint #8
Substitute 2×C for D (from eq.2d) in eq.2c: 2×2×C - 3×C = 2×B which is equivalent to 4×C - 3×C = 2×B which makes C = 2×B
Hint #9
Substitute 2×B for C in eq.4a: B + 2×B = E which makes 3×B = E
Hint #10
Substitute (2×B) for C in eq.2d: 2×(2×B) = D which means 4×B = D
Hint #11
Substitute (3×B) for E in eq.5a: 1½×(3×B) = A which means 4½×B = A
Hint #12
Substitute 4×B for D, and 2×B for C in eq.3a: F = 4×B - 2×B which makes F = 2×B
Solution
Substitute 4½×B for A, 2×B for C and F, 4×B for D, and 3×B for E in eq.1: 4½×B + B + 2×B + 4×B + 3×B + 2×B = 33 which simplifies to 16½×A = 33 Divide both sides by 16½: 16½×A ÷ 16½ = 33 ÷ 16½ which makes A = 2 making A = 4½×B = 4½ × 2 = 9 C = F = 2×B = 2 × 2 = 4 D = 4×B = 4 × 2 = 8 E = 3×B = 3 × 2 = 6 and ABCDEF = 924864