Puzzle for September 30, 2023 ( )
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).
Scratchpad
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Hint #1
Add B and E to both sides of eq.2: D - B + B + E = A + B - E + B + E which becomes eq.2a) D + E = A + 2×B
Hint #2
Add A and B to both sides of eq.3: B - C + F + A + B = D - A - B + A + B which becomes 2×B - C + F + A = D which may be written as eq.3a) A + 2×B - C + F = D
Hint #3
In eq.3a, replace A + 2×B with D + E (from eq.2a): D + E - C + F = D In the equation above, subtract D from both sides, and add C to both sides: D + E - C + F - D + C = D - D + C which becomes eq.3b) E + F = C
Hint #4
Add B and F to both sides of eq.4: E + F - B + B + F = A + B - C - F + B + F which becomes eq.4a) E + 2×F = A + 2×B - C
Hint #5
In eq 3a, replace A + 2×B - C with E + 2×F (from eq.4a): E + 2×F + F = D which becomes eq.3c) E + 3×F = D
Hint #6
eq.6 may be written as: eq.6a) 10×D + E = C + 10×E + F
Hint #7
In eq.6a, substitute (E + 3×F) for D (from eq.3c), and E + F for C (from eq.3b): 10×(E + 3×F) + E = E + F + 10×E + F which becomes 10×E + 30×F + E = 11×E + 2×F which becomes 11×E + 30×F = 11×E + 2×F Subtract 11×E and 2×F from each side of the equation above: 11×E + 30×F - 11×E - 2×F = 11×E + 2×F - 11×E - 2×F which makes 28×F = 0 which means F = 0
Hint #8
Substitute 0 for F in eq.3b: E + 0 = C which makes E = C
Hint #9
Substitute 0 for F in eq.3c: E + 3×0 = D which becomes E + 0 = D which makes E = D
Hint #10
eq.5 may be written as: eq.5a) 10×A + B - F = 10×C + D - B - E
Hint #11
Substitute 0 for F, and C for D and E in eq.5a: 10×A + B - 0 = 10×C + C - B - C which becomes 10×A + B = 10×C - B In the above equation, add B to both sides, and subtract 10×A from both sides: 10×A + B + B - 10×A = 10×C - B + B - 10×A which becomes eq.5b) 2×B = 10×C - 10×A
Hint #12
Substitute C for D and E, and 10×C - 10×A for 2×B (from eq.5b) into eq.2a: C + C = A + 10×C - 10×A which becomes 2×C = 10×C - 9×A Subtract 10×C from both sides of the equation above: 2×C - 10×C = 10×C - 9×A - 10×C which makes -8×C = -9×A Divide both sides by (-8): -8×C ÷ (-8) = -9×A ÷ (-8) which makes C = 1⅛×A and also makes C = D = E = 1⅛×A
Hint #13
Substitute (1⅛×A) for C in eq.5b: 2×B = 10×(1⅛×A) - 10×A which becomes 2×B = 11¼×A - 10×A which becomes 2×B = 1¼×A Divide both sides of the above equation by 2: 2×B ÷ 2 = 1¼×A ÷ 2 which becomes B = ⅝×A
Solution
Substitute ⅝×A for B, 1⅛×A for C and D and E, and 0 for F in eq.1: A + ⅝×A + 1⅛×A + 1⅛×A + 1⅛×A + 0 = 40 which simplifies to 5×A = 40 Divide both sides of the above equation by 5: 5×A ÷ 5 = 40 ÷ 5 which means A = 8 making B = ⅝×A = ⅝ × 8 = 5 C = D = E = 1⅛×A = 1⅛ × 8 = 9 and ABCDEF = 859990