Puzzle for January 15, 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
Add E and D to both sides of eq.2: B – E + E + D = A – D + E + D which becomes eq.2a) B + D = A + E Add 2×D to both sides of eq.4: D + E + 2×D = B + F – D + 2×D which becomes 3×D + E = B + F + D which may be written as eq.4a) 3×D + E = B + D + F
Hint #2
In eq.4a, replace B + D with A + E (from eq.2a): 3×D + E = A + E + F Subtract E and A from both sides of the above equation: 3×D + E – E – A = A + E + F – E – A which becomes eq.4b) 3×D – A = F
Hint #3
In eq.3, replace F with 3×D – A (from eq.4b): 3×D – A – A = A – D which becomes 3×D – 2×A = A – D Add 2×A and D to both sides of the above equation: 3×D – 2×A + 2×A + D = A – D + 2×A + D which makes 4×D = 3×A Divide both sides by 4: 4×D ÷ 4 = 3×A ÷ 4 which makes D = ¾×A
Hint #4
In eq.4b, substitute (¾×A) for D: 3×(¾×A) – A = F which becomes 2¼×A – A = F which makes 1¼×A = F
Hint #5
Substitute ¾×A for D in eq.2: B – E = A – ¾×A which becomes B – E = ¼×A Add E to both sides of the above equation: B – E + E = ¼×A + E which becomes eq.4c) B = ¼×A + E
Hint #6
eq.6 may be written as: C = (A + D + E) ÷ 3 Substitute ¾×A for D in the equation above: C = (A + ¾×A + E) ÷ 3 which becomes eq.6a) C = (1¾×A + E) ÷ 3 Multiply both sides of eq.3a by 3: 3 × C = 3 × (1¾×A + E) ÷ 3 which becomes eq.6b) 3×C = 1¾×A + E
Hint #7
Add C to both sides of eq.5: B + E – C + C = A + D + C + C which becomes B + E = A + D + 2×C Multiply both sides of the above equation by 3: 3×(B + E) = 3×(A + D + 2×C) which becomes 3×B + 3×E = 3×A + 3×D + 3×(2×C) which is equivalent to eq.5a) 3×B + 3×E = 3×A + 3×D + 2×(3×C)
Hint #8
Substitute (¼×A + E) for B (from eq.4c), (¾×A) for D, and 1¾×A + E for 3×C (from eq.6b) in eq.5a: 3×(¼×A + E) + 3×E = 3×A + 3×(¾×A) + 2×(1¾×A + E) which becomes ¾×A + 3×E + 3×E = 3×A + 2¼×A + 3½×A + 2×E which becomes ¾×A + 6×E = 8¾×A + 2×E Subtract ¾×A and 2×E from each side of the above equation: ¾×A + 6×E – ¾×A – 2×E = 8¾×A + 2×E – ¾×A – 2×E which simplifies to 4×E = 8×A Divide both sides by 4: 4×E ÷ 4 = 8×A ÷ 4 which makes E = 2×A
Hint #9
Substitute 2×A for E in eq.4c: B = ¼×A + 2×A which makes B = 2¼×A
Hint #10
Substitute 2×A for E in eq.6a: C = (1¾×A + 2×A) ÷ 3 which becomes C = 3¾×A ÷ 3 which makes C = 1¼×A
Solution
Substitute 2¼×A for B, 1¼×A for C and F, ¾×A for D, and 2×A for E in eq.1: A + 2¼×A + 1¼×A + ¾×A + 2×A + 1¼×A = 34 which simplifies to 8½×A = 34 Divide both sides of the above equation by 8½: 8½×A ÷ 8½ = 34 ÷ 8½ which means A = 4 making B = 2¼×A = 2¼ × 4 = 9 C = F = 1¼×A = 1¼ × 4 = 5 D = ¾×A = ¾ × 4 = 3 E = 2×A = 2 × 4 = 8 and ABCDEF = 495385