Puzzle for July 25, 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.
* AB, DE, and EF are 2-digit numbers (not A×B, D×E, or E×F).
Scratchpad
Help Area
Hint #1
Add B, E, and C to both sides of eq.3: E – C + B + E + C = A – B – E + B + E + C which becomes eq.3a) B + 2×E = A + C Add D and E to both sides of eq.2: B + D + E + D + E = A – D – E + F + D + E which becomes B + 2×D + 2×E = A + F which may be written as eq.2a) B + 2×E + 2×D = A + F
Hint #2
In eq.2a, replace B + 2×E with A + C (from eq.3a): A + C + 2×D = A + F Subtract A from both sides of the above equation: A + C + 2×D – A = A + F – A which becomes eq.2b) C + 2×D = F
Hint #3
Add D to both sides of eq.4: F + C – D + D = D + D which becomes eq.4a) F + C = 2×D In eq.2b, replace 2×D with F + C (from eq.4a): C + F + C = F which becomes 2×C + F = F Subtract F from both sides of the equation above: 2×C + F – F = F – F which makes 2×C = 0 which means C = 0
Hint #4
In eq.2b, substitute 0 for C: 0 + 2×D = F which makes 2×D = F
Hint #5
eq.5 may be written as: 10×D + E + E = 10×E + F + F which becomes 10×D + 2×E = 10×E + 2×F Subtract 2×E from each side of the equation above: 10×D + 2×E – 2×E = 10×E + 2×F – 2×E which becomes 10×D = 8×E + 2×F Divide each side by 2: 10×D ÷ 2 = (8×E + 2×F) ÷ 2 which becomes eq.5a) 5×D = 4×E + F
Hint #6
Substitute 2×D for F in eq.5a: 5×D = 4×E + 2×D Subtract 2×D from each side of the above equation: 5×D – 2×D = 4×E + 2×D – 2×D which makes 3×D = 4×E Divide both sides by 4: 3×D ÷ 4 = 4×E ÷ 4 which makes ¾×D = E
Hint #7
eq.6 may be written as: 10×A + B = 10×D + E + A + B + E + F Subtract A and B from both sides of the equation above: 10×A + B – A – B = 10×D + E + A + B + E + F – A – B which becomes 9×A = 10×D + 2×E + F Substitute (¾×D) for E, and 2×D for F: 9×A = 10×D + 2×(¾×D) + 2×D which is equivalent to 9×A = 12×D + 1½×D which becomes 9×A = 13½×D Divide both sides by 9: 9×A ÷ 9 = 13½×D ÷ 9 which makes A = 1½×D
Hint #8
Substitute (¾×D) for E, 1½×D for A, and 0 for C in eq.3a: B + 2×(¾×D) = 1½×D + 0 which becomes B + 1½×D = 1½×D Subtract 1½×D from each side of the above equation: B + 1½×D – 1½×D = 1½×D – 1½×D which makes B = 0
Solution
Substitute 1½×D for A, 0 for B and C, ¾×D for E, and 2×D for F in eq.1: 1½×D + 0 + 0 + D + ¾×D + 2×D = 21 which simplifies to 5¼×D = 21 Divide both sides of the equation above by 5¼: 5¼×D ÷ 5¼ = 21 ÷ 5¼ which means D = 4 making A = 1½×D = 1½ × 4 = 6 E = ¾×D = ¾ × 4 = 3 F = 2×D = 2 × 4 = 8 and ABCDEF = 600438