Puzzle for January 11, 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, CD, and EF are 2-digit numbers (not A×B, C×D, or E×F).
Scratchpad
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Hint #1
Add D to both sides of eq.5: D + E + F + D = A – D + D which becomes eq.5a) 2×D + E + F = A In eq.2, replace A with 2×D + E + F (from eq.5a): 2×D + E + F + B = C + D + E Subtract both D and E from each side of the equation above: 2×D + E + F + B – D – E = C + D + E – D – E which becomes eq.2a) D + F + B = C
Hint #2
In eq.3, replace C with D + F + B (from eq.2a): D + F + B = D + F Subtract both D and F from each side: D + F + B – D – F = D + F – D – F which makes B = 0
Hint #3
eq.5 may be written as: D + F + E = A – D In the above equation, substitute C for D + F (from eq.3): C + E = A – D Subtract E from each side of the above equation: C + E – E = A – D – E which becomes eq.5b) C = A – D – E
Hint #4
Substitute A – D – E for C (from eq.5b), and 0 for B in eq.4: A – D – E – E = A – 0 – F which becomes A – D – 2×E = A – F Subtract A from each side of the equation above: A – D – 2×E – A = A – F – A which becomes –D – 2×E = –F Add D + F to each side: –D – 2×E + D + F = –F + D + F which becomes eq.4a) F – 2×E = D
Hint #5
Substitute F – 2×E for D (from eq.4a) in eq.3: C = F – 2×E + F which becomes eq.3a) C = 2×F – 2×E
Hint #6
eq.6 may be written as: 10×A + B – C + E = 10×C + D + 10×E + F In the above equation, add C to each side, and subtract E from each side: 10×A + B – C + E + C – E = 10×C + D + 10×E + F + C – E which becomes eq.6a) 10×A + B = 11×C + D + 9×E + F
Hint #7
Substitute (2×D + E + F) for A (from eq.5a), and 0 for B in eq.6a: 10×(2×D + E + F) + 0 = 11×C + D + 9×E + F which may be written as 20×D + 10×E + 10×F = 11×C + D + 9×E + F Subtract D, 9×E, and F from both sides of the equation above: 20×D + 10×E + 10×F – D – 9×E – F = 11×C + D + 9×E + F – D – 9×E – F which simplifies to eq.6b) 19×D + E + 9×F = 11×C
Hint #8
Substitute (F – 2×E) for D (from eq.4a), and (2×F – 2×E) for C (from eq.3a) in eq.6b: 19×(F – 2×E) + E + 9×F = 11×(2×F – 2×E) which is the same as 19×F – 38×E + E + 9×F = 22×F – 22×E which becomes 28×F – 37×E = 22×F – 22×E In the above equation, add 37×E to each side, and subtract 22×F from each side: 28×F – 37×E + 37×E – 22×F = 22×F – 22×E + 37×E – 22×F which simplifies to 6×F = 15×E Divide both sides by 6: 6×F ÷ 6 = 15×E ÷ 6 which makes F = 2½×E
Hint #9
Substitute (2½×E) for F in eq.3a: C = 2×(2½×E) – 2×E which is equivalent to C = 5×E – 2×E which means C = 3×E
Hint #10
Substitute 2½×E for F in eq.4a: 2½×E – 2×E = D which makes ½×E = D
Hint #11
Substitute (½×E) for D, and 2½×E for F in eq.5a: 2×(½×E) + E + 2½×E = A which makes 4½×E = A
Solution
Substitute 4½×E for A, 0 for B, 3×E for C, ½×E for D, and 2½×E for F in eq.1: 4½×E + 0 + 3×E + ½×E + E + 2½×E = 23 which simplifies to 11½×E = 23 Divide each side of the equation above by 11½: 11½×E ÷ 11½ = 23 ÷ 11½ which means E = 2 making A = 4½×E = 4½ × 2 = 9 C = 3×E = 3×2 = 6 D = ½×E = ½ × 2 = 1 F = 2½×E = 2½ × 2 = 5 and ABCDEF = 906125