Puzzle for October 7, 2022 ( )
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.
Scratchpad
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Hint #1
In eq.3, replace D + E with C + F (from eq.2): C + F - C = A + B which becomes eq.3a) F = A + B
Hint #2
eq.5 may be written as: B = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (C + D + F) ÷ 3 which becomes eq.5a) 3×B = C + D + F
Hint #3
eq.6 may be written as: A = (B + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + C + D) ÷ 3 which becomes 3×A = B + C + D Subtract B from both sides: 3×A - B = B + C + D - B which becomes eq.6a) 3×A - B = C + D
Hint #4
In eq.5a, replace C + D with 3×A - B (from eq.6a): 3×B = 3×A - B + F Add B to both sides of the above equation: 3×B + B = 3×A - B + F + B which becomes eq.5b) 4×B = 3×A + F
Hint #5
In eq.5b, substitute A + B for F (from eq.3a): 4×B = 3×A + A + B which makes 4×B = 4×A + B Subtract B from each side of the equation above: 4×B - B = 4×A + B - B which becomes 3×B = 4×A Divide both sides by 4: 3×B ÷ 4 = 4×A ÷ 4 which makes ¾×B = A
Hint #6
Substitute ¾×B for A in eq.3a: F = ¾×B + B which makes F = 1¾×B
Hint #7
Substitute ¾×B for A, and 1¾×B for F in eq.4: C + D - ¾×B = 1¾×B - D In the above equation, subtract D from both sides, and add ¾×B to both sides: C + D - ¾×B - D + ¾×B = 1¾×B - D - D + ¾×B which becomes eq.4a) C = 2½×B - 2×D
Hint #8
Substitute 2½×B - 2×D for C (from eq.4a), and 1¾×B for F in eq.5a: 3×B = 2½×B - 2×D + D + 1¾×B which becomes 3×B = 4¼×B - D In the equation above, add D to both sides, and subtract 3×B from both sides: 3×B + D - 3×B = 4¼×B - D + D - 3×B which becomes D = 1¼×B
Hint #9
Substitute (1¼×B) for D in eq.4a: C = 2½×B - 2×(1¼×B) which becomes C = 2½×B - 2½×B which makes C = 0
Hint #10
Substitute 0 for C, 1¾×B for F, and 1¼×B for D in eq.2: 0 + 1¾×B = 1¼×B + E which becomes 1¾×B = 1¼×B + E Subtract 1¼×B from both sides of the above equation: 1¾×B - 1¼×B = 1¼×B + E - 1¼×B which makes ½×B = E
Solution
Substitute ¾×B for A, 0 for C, 1¼×B for D, ½×B for E, and 1¾×B for F in eq.1: ¾×B + B + 0 + 1¼×B + ½×B + 1¾×B = 21 which simplifies to 5¼×B = 21 Divide both sides of the above equation by 5¼: 5¼×B ÷ 5¼ = 21 ÷ 5¼ which means B = 4 making A = ¾×B = ¾ × 4 = 3 D = 1¼×B = 1¼ × 4 = 5 E = ½×B = ½ × 4 = 2 F = 1¾×B = 1¾ × 4 = 7 and ABCDEF = 340527