Puzzle for May 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 is a 2-digit number (not B×C).
Scratchpad
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Hint #1
In eq.2, replace B + C with A + E (from eq.1): D + E = A + A + E which becomes D + E = 2×A + E Subtract E from each side of the equation above: D + E – E = 2×A + E – E which becomes eq.2a) D = 2×A
Hint #2
Add D to both sides of eq.3: F – D + D = C + D + D which becomes F = C + 2×D In the above equation, replace D with (2×A): F = C + 2×(2×A) which becomes eq.3a) F = C + 4×A
Hint #3
eq.5 may be written as: eq.5a) 10×B + C = A + F In eq.5a, substitute C + 4×A for F (from eq.3a): 10×B + C = A + C + 4×A which becomes 10×B + C = C + 5×A Subtract C from both sides of the equation above: 10×B + C – C = C + 5×A – C which makes 10×B = 5×A Divide both sides by 5: 10×B ÷ 5 = 5×A ÷ 5 which means 2×B = A
Hint #4
In eq.2a, substitute (2×B) for A: D = 2×(2×B) which makes D = 4×B
Hint #5
Substitute 2×B for A, and 4×B for D in eq.4: 2×B + B + 4×B = E + F – 2×B which becomes 7×B = E + F – 2×B Add 2×B to both sides of the equation above: 7×B + 2×B = E + F – 2×B + 2×B which becomes eq.4a) 9×B = E + F
Hint #6
Substitute 2×B for A in eq.5a: 10×B + C = 2×B + F Subtract 2×B from each side of the above equation: 10×B + C – 2×B = 2×B + F – 2×B which becomes eq.5b) 8×B + C = F
Hint #7
Substitute 8×B + C for F (from eq.5b) in eq.4a: 9×B = E + 8×B + C Subtract 8×B from each side of the equation above: 9×B – 8×B = E + 8×B + C – 8×B which becomes eq.4b) B = E + C
Hint #8
Substitute E + C for B (from eq.4b), and 2×B for A in eq.1: E + C + C = 2×B + E Subtract E from both sides of the equation above: E + C + C – E = 2×B + E – E which makes 2×C = 2×B Divide both sides by 2: 2×C ÷ 2 = 2×B ÷ 2 which makes C = B
Hint #9
Substitute B for C in eq.4b: B = E + B Subtract B from each side of the above equation: B – B = E + B – B which means 0 = E
Hint #10
Substitute 0 for E in eq.4a: 9×B = 0 + F which makes 9×B = F
Solution
Substitute 0 for E, 9×B for F, 2×B for A, 4×B for D, and B for C in eq.6: 0 + (9×B ÷ B) = (B + (2×B × 4×B)) ÷ B which becomes 0 + 9 = (B + 8×B²) ÷ B which becomes 9 = 1 + 8×B Subtract 1 from each side of the equation above: 9 – 1 = 1 + 8×B – 1 which makes 8 = 8×B Divide both sides by 8: 8 ÷ 8 = 8×B ÷ 8 which means 1 = B making A = 2×B = 2 × 1 = 2 C = B = 1 D = 4×B = 4 × 1 = 4 F = 9×B = 9 × 1 = 9 and ABCDEF = 211409