Puzzle for December 18, 2021 ( )
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
Add B, D, E, and F to both sides of eq.5: C – B + D – E + B + D + E + F = B – D + E – F + B + D + E + F which becomes C + 2×D + F = 2×B + 2×E which is the same as eq.5a) C + 2×D + F = 2×(B + E)
Hint #2
Add A and C to both sides of eq.3: C – A + A + C = A + B – C + A + C which becomes eq.3a) 2×C = 2×A + B In eq.4, add A and F to both sides, and subtract D from both sides: E – A + A + F – D = A + D – F + A + F – D which becomes eq.4a) E + F – D = 2×A
Hint #3
In eq.3a, replace 2×A with E + F – D (from eq.4a): 2×C = E + F – D + B In the equation above, subtract F from both sides, and add D to both sides: 2×C – F + D = E + F – D + B – F + D which becomes 2×C – F + D = E + B which may be written as eq.3b) 2×C – F + D = B + E
Hint #4
In eq.5a, replace B + E with 2×C – F + D (from eq.3b): C + 2×D + F = 2×(2×C – F + D) which becomes C + 2×D + F = 4×C – 2×F + 2×D In the above equation, subtract C and 2×D from both sides, and add 2×F to both sides: C + 2×D + F – C – 2×D + 2×F = 4×C – 2×F + 2×D – C – 2×D + 2×F which simplifies to 3×F = 3×C Divide both sides by 3: 3×F ÷ 3 = 3×C ÷ 3 which makes F = C
Hint #5
In eq.2, substitute C for F: B + C = C + D Subtract C from each side of the equation above: B + C – C = C + D – C which makes B = D
Hint #6
Substitute B for D, and C for F in eq.5a: C + 2×B + C = 2×(B + E) which becomes 2×C + 2×B = 2×B + 2×E Subtract 2×B from each side of the above equation: 2×C + 2×B – 2×B = 2×B + 2×E – 2×B which becomes 2×C = 2×E Divide both sides by 2: 2×C ÷ 2 = 2×E ÷ 2 which makes C = E
Hint #7
eq.6 may be written as: 10×B + C = C + D + E + F Subtract C from each side of the equation above: 10×B + C – C = C + D + E + F – C which makes eq.6a) 10×B = D + E + F
Hint #8
In eq.6a, substitute B for D, and C for F and E: 10×B = B + C + C which becomes 10×B = B + 2×C Subtract B from both sides of the equation above: 10×B – B = B + 2×C – B which makes 9×B = 2×C Divide both sides by 2: 9×B ÷ 2 = 2×C ÷ 2 which makes 4½×B = C and also makes E = F = C = 4½×B
Hint #9
Substitute (4½×B) for C in eq.3a: 2×(4½×B) = 2×A + B which becomes 9×B = 2×A + B Subtract B from both sides of the above equation: 9×B – B = 2×A + B – B which makes 8×B = 2×A Divide both sides by 2: 8×B ÷ 2 = 2×A ÷ 2 which makes 4×B = A
Solution
Substitute 4×B for A, 4½×B for C and E and F, and B for D in eq.1: 4×B + B + 4½×B + B + 4½×B + 4½×B = 39 which simplifies to 19½×B = 39 Divide both sides of the above equation by 19½: 19½×B ÷ 19½ = 39 ÷ 19½ which means B = 2 making A = 4×B = 4 × 2 = 8 C = E = F = 4½×B = 4½ × 2 = 9 D = B = 2 and ABCDEF = 829299