Puzzle for September 24, 2020 ( )
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 and DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
In eq.2, add B and D to both sides: D – B + B + D = A – D + B + D which becomes eq.2a) 2×D = A + B Add D, C, and E to both sides of eq.4: A – D + E + D + C + E = D – C – E + D + C + E which simplifies to eq.4a) A + 2×E + C = 2×D
Hint #2
In eq.4a, replace 2×D with A + B (from eq.2a): A + 2×E + C = A + B Subtract A from each side of the equation above: A + 2×E + C – A = A + B – A which becomes eq.4b) 2×E + C = B
Hint #3
In eq.3, substitute 2×E + C for B (from eq.4b): C – F = 2×E + C – E which becomes C – F = E + C Subtract C from both sides of the above equation: C – F – C = E + C – C which makes –F = E Since E and F must be non-negative integers, the above equation makes: F = 0 and E = 0
Hint #4
Substitute 0 for E in eq.4b: 2×0 + C = B which makes C = B
Hint #5
eq.5 may be written as: B + C + 10×B + C = 10×D + E – C which becomes 11×B + 2×C = 10×D + E – C Add C to each side of the above equation: 11×B + 2×C + C = 10×D + E – C + C which becomes eq.5a) 11×B + 3×C = 10×D + E
Hint #6
Substitute B for C, and 0 for E in eq.5a: 11×B + 3×B = 10×D + 0 which means 14×B = 10×D Divide both sides of the equation above by 10: 14×B ÷ 10 = 10×D ÷ 10 which makes 1⅖×B = D
Hint #7
Substitute (1⅖×B) for D in eq.2a: 2×(1⅖×B) = A + B which becomes 2⅘×B = A + B Subtract B from each side of the above equation: 2⅘×B – B = A + B – B which makes 1⅘×B = A
Solution
Substitute 1⅘×B for A, B for C, 1⅖×B for D, and 0 for E and F in eq.1: 1⅘×B + B + B + 1⅖×B + 0 + 0 = 26 which simplifies to 5⅕×B = 26 Divide both sides of the equation above by 5⅕: 5⅕×B ÷ 5⅕ = 26 ÷ 5⅕ which means B = 5 making A = 1⅘×B = 1⅘ × 5 = 9 C = B = 5 D = 1⅖×B = 1⅖ × 5 = 7 and ABCDEF = 955700