Puzzle for December 12, 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.
* EF and DE are 2-digit numbers (not E×F or D×E).
Scratchpad
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Hint #1
Add B to both sides of eq.2: B + E + B = C + D – B + B which becomes eq.2a) 2×B + E = C + D Add D to both sides of eq.4: D + E – F + D = B + C + F – E + D which becomes eq.4a) 2×D + E – F = B + C + D + F – E
Hint #2
In eq.4a, replace C + D with 2×B + E (from eq.2a): 2×D + E – F = B + 2×B + E + F – E which becomes 2×D + E – F = 3×B + F Add F to both sides of the above equation: 2×D + E – F + F = 3×B + F + F which becomes eq.4b) 2×D + E = 3×B + 2×F
Hint #3
eq.6 may be written as: F = (B + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (B + D + E) ÷ 3 which becomes 3×F = B + D + E Subtract B and D from each side of the above equation: 3×F – B – D = B + D + E – B – D which becomes eq.6a) 3×F – B – D = E
Hint #4
In eq.4b, replace E with 3×F – B – D (from eq.6a): 2×D + 3×F – B – D = 3×B + 2×F which becomes D + 3×F – B = 3×B + 2×F In the above equation, add B to both sides, and subtract D and 2×F from both sides: D + 3×F – B + B – D – 2×F = 3×B + 2×F + B – D – 2×F which simplifies to eq.4c) F = 4×B – D
Hint #5
In eq.6a, substitute (4×B – D) for F (from eq.4c): 3×(4×B – D) – B – D = E which becomes 12×B – 3×D – B – D = E which becomes eq.6b) 11×B – 4×D = E
Hint #6
Add D and A to both sides of eq.3: A + E – D + D + A = B + D – A + D + A which becomes eq.3a) 2×A + E = B + 2×D eq.5 may be written as: A + B + 10×E + F = 10×D + E Subtract B, 10×E, and F from each side of the equation above: A + B + 10×E + F – B – 10×E – F = 10×D + E – B – 10×E – F which becomes eq.5a) A = 10×D – B – 9×E – F
Hint #7
Substitute 10×D – B – 9×E – F for A (from eq.5a) into eq.3a: 2×(10×D – B – 9×E – F) + E = B + 2×D which becomes 20×D – 2×B – 18×E – 2×F + E = B + 2×D which becomes 20×D – 2×B – 17×E – 2×F = B + 2×D In the above equation, add 2×B to both sides, and subtract 2×D from both sides: 20×D – 2×B – 17×E – 2×F + 2×B – 2×D = B + 2×D + 2×B – 2×D which becomes eq.3b) 18×D – 17×E – 2×F = 3×B
Hint #8
Substitute (11×B – 4×D) for E (from eq.6b), and (4×B – D) for F (from eq.4c) in eq.3b: 18×D – 17×(11×B – 4×D) – 2×(4×B – D) = 3×B which becomes 18×D – 187×B + 68×D – 8×B + 2×D = 3×B which becomes 88×D – 195×B = 3×B Add 195×B to both sides of the above equation: 88×D – 195×B + 195×B = 3×B + 195×B which makes 88×D = 198×B Divide both sides by 88: 88×D ÷ 88 = 198×B ÷ 88 which makes D = 2¼×B
Hint #9
Substitute 2¼×B for D in eq.4c: F = 4×B – 2¼×B which makes F = 1¾×B
Hint #10
Substitute (2¼×B) for D in eq.6b: 11×B – 4×(2¼×B) = E which becomes 11×B – 9×B = E which makes 2×B = E
Hint #11
Substitute (2¼×B) for D, (2×B) for E, and 1¾×B for F in eq.5a: A = 10×(2¼×B) – B – 9×(2×B) – 1¾×B which becomes A = 22½×B – B – 18×B – 1¾×B which makes A = 1¾×B
Hint #12
Substitute 2×B for E, and 2¼×B for D in eq.2a: 2×B + 2×B = C + 2¼×B which becomes 4×B = C + 2¼×B Subtract 2¼×B from each side of the equation above: 4×B – 2¼×B = C + 2¼×B – 2¼×B which makes 1¾×B = C
Solution
Substitute 1¾×B for A and C and F, 2¼×B for D, and 2×B for E in eq.1: 1¾×B + B + 1¾×B + 2¼×B + 2×B + 1¾×B = 42 which simplifies to 10½×B = 42 Divide both sides of the above equation by 10½: 10½×B ÷ 10½ = 42 ÷ 10½ which means B = 4 making A = C = F = 1¾×B = 1¾ × 4 = 7 D = 2¼×B = 2¼ × 4 = 9 E = 2×B = 2 × 4 = 8 and ABCDEF = 747987