Puzzle for March 15, 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.
* DE, EF, and AB are 2-digit numbers (not D×E, E×F, or A×B).
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Hint #1
eq.5 may be written as: eq.5a) A + B + F = 10×D + E Add E and A to both sides of eq.4: B - E + E + A = D - A + E + A which becomes B + A = D + E which may be written as eq.4a) A + B = D + E
Hint #2
In eq.5a, replace A + B with D + E (from eq.4a): D + E + F = 10×D + E Subtract D and E from each side of the equation above: D + E + F - D - E = 10×D + E - D - E which simplifies to F = 9×D
Hint #3
In eq.3, replace F with 9×D: A - D = 9×D - A Add D and A to both sides of the above equation: A - D + D + A = 9×D - A + D + A which makes 2×A = 10×D Divide both sides by 2: 2×A ÷ 2 = 10×D ÷ 2 which makes A = 5×D
Hint #4
In eq.2, substitute 9×D for F, and 5×D for A: 9×D = 5×D + E Subtract 5×D from both sides of the equation above: 9×D - 5×D = 5×D + E - 5×D which makes 4×D = E
Hint #5
eq.6 may be written as: D + 10×E + F = 10×A + B + C Substitute (4×D) for E, 9×D for F, and (5×D) for A in the above equation: D + 10×(4×D) + 9×D = 10×(5×D) + B + C which becomes D + 40×D + 9×D = 50×D + B + C which becomes 50×D = 50×D + B + C Subtract 50×D from each side: 50×D - 50×D = 50×D + B + C - 50×D which makes 0 = B + C Since B and C must be non-negative, the above equation makes: B = 0 and C = 0
Solution
Substitute 5×D for A, 0 for B and C, 4×D for E, and 9×D for F in eq.1: 5×D + 0 + 0 + D + 4×D + 9×D = 19 which simplifies to 19×D = 19 Divide both sides of the above equation by 19: 19×D ÷ 19 = 19 ÷ 19 which means D = 1 making A = 5×D = 5 × 1 = 5 E = 4×D = 4 × 1 = 4 F = 9×D = 9 × 1 = 9 and ABCDEF = 500149