Puzzle for June 28, 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 EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
Subtract D from both sides of eq.3: A + E – D = B + C + D – D which becomes eq.3a) A + E – D = B + C In eq.2, replace B + C with A + E – D (from eq.3a): A + E – D = D + E + F In the equation above, subtract E from both sides, and add D to both sides: A + E – D – E + D = D + E + F – E + D which becomes eq.2a) A = 2×D + F
Hint #2
In eq.5, replace A with 2×D + F (from eq.2a): E – D = 2×D + F – E – F which becomes E – D = 2×D – E Add D and E to both sides of the above equation: E – D + D + E = 2×D – E + D + E which makes 2×E = 3×D Divide both sides by 2: 2×E ÷ 2 = 3×D ÷ 2 which becomes E = 1½×D
Hint #3
eq.6 may be written as: C + D + E = 10×B + C – (10×E + F) which is the same as C + D + E = 10×B + C – 10×E – F Subtract C from both sides of the above equation: C + D + E – C = 10×B + C – 10×E – F – C which becomes eq.6a) D + E = 10×B – 10×E – F
Hint #4
In eq.2, replace D + E with 10×B – 10×E – F (from eq.6a): B + C = 10×B – 10×E – F + F which becomes B + C = 10×B – 10×E Subtract B from each side of the above equation: B + C – B = 10×B – 10×E – B which becomes eq.2b) C = 9×B – 10×E
Hint #5
Substitute (1½×D) for E in eq.2b: C = 9×B – 10×(1½×D) which becomes eq.2c) C = 9×B – 15×D
Hint #6
Add 10×E and F to both sides of eq.6a: D + E + 10×E + F = 10×B – 10×E – F + 10×E + F which becomes D + 11×E + F = 10×B Substitute (1½×D) for E in the equation above: D + 11×(1½×D) + F = 10×B which becomes D + 16½×D + F = 10×B which becomes 17½×D + F = 10×B Subtract 17½×D from both sides: 17½×D + F – 17½×D = 10×B – 17½×D which becomes eq.6b) F = 10×B – 17½×D
Hint #7
Substitute 10×B – 17½×D for F in eq.2a: A = 2×D + 10×B – 17½×D which becomes eq.2d) A = 10×B – 15½×D
Hint #8
Substitute (9×B – 15×D) for C (from eq.2c), 10×B – 15½×D for A (from eq.2d), and 10×B – 17½×D for F (from eq.6b) in eq.4: D – (9×B – 15×D) = 10×B – 15½×D – B + 10×B – 17½×D which becomes D – 9×B + 15×D = 19×B – 33×D which becomes 16×D – 9×B = 19×B – 33×D Add 9×B and 33×D to both sides of the equation above: 16×D – 9×B + 9×B + 33×D = 19×B – 33×D + 9×B + 33×D which makes 49×D = 28×B Divide both sides by 28: 49×D ÷ 28 = 28×B ÷ 28 which makes 1¾×D = B
Hint #9
Substitute (1¾×D) for B in eq.2d: A = 10×(1¾×D) – 15½×D which becomes A = 17½×D – 15½×D which makes A = 2×D
Hint #10
Substitute (1¾×D) for B in eq.2c: C = 9×(1¾×D) – 15×D which becomes C = 15¾×D – 15×D which makes C = ¾×D
Hint #11
Substitute (1¾×D) for B in eq.6b: F = 10×(1¾×D) – 17½×D which becomes F = 17½×D – 17½×D which makes F = 0
Solution
Substitute 2×D for A, 1¾×D for B, ¾×D for C, 1½×D for E, and 0 for F in eq.1: 2×D + 1¾×D + ¾×D + D + 1½×D + 0 = 28 which simplifies to 7×D = 28 Divide both sides of the equation above by 7: 7×D ÷ 7 = 28 ÷ 7 which means D = 4 making A = 2×D = 2 × 4 = 8 B = 1¾×D = 1¾ × 4 = 7 C = ¾×D = ¾ × 4 = 3 E = 1½×D = 1½ × 4 = 6 and ABCDEF = 873460