Puzzle for October 15, 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.
Scratchpad
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Hint #1
Add B and D to both sides of eq.3: D – B + B + D = F – D + B + D which becomes eq.3a) 2×D = F + B Add A, B, and D to both sides of eq.5: B – A + A + B + D = A – B – D + F + A + B + D which becomes 2×B + D = 2×A + F Multiply both sides of the above equation by 2: (2×B + D) × 2 = (2×A + F) × 2 which becomes eq.5a) 4×B + 2×D = 4×A + 2×F
Hint #2
In eq.5a, replace 2×D with F + B (from eq.3a): 4×B + F + B = 4×A + 2×F which becomes 5×B + F = 4×A + 2×F Subtract F from each side of the equation above: 5×B + F – F = 4×A + 2×F – F which becomes eq.5b) 5×B = 4×A + F
Hint #3
In eq.5b, replace F with A + B (from eq.2): 5×B = 4×A + A + B which becomes 5×B = 5×A + B Subtract B from each side of the above equation: 5×B – B = 5×A + B – B which becomes 4×B = 5×A Divide both sides by 4: 4×B ÷ 4 = 5×A ÷ 4 which makes B = 1¼×A
Hint #4
In eq.5b, substitute (1¼×A) for B: 5×(1¼×A) = 4×A + F which becomes 6¼×A = 4×A + F Subtract 4×A from both sides of the equation above: 6¼×A – 4×A = 4×A + F – 4×A which makes 2¼×A = F
Hint #5
Substitute 2¼×A for F, and 1¼×A for B in eq.3a: 2×D = 2¼×A + 1¼×A which becomes 2×D = 3½×A Divide both sides of the above equation by 2: 2×D ÷ 2 = 3½×A ÷ 2 which makes D = 1¾×A
Hint #6
Substitute 1¾×A for D, 1¼×A for B, and 2¼×A for F in eq.4: C + 1¾×A = 1¼×A + E + 2¼×A which becomes C + 1¾×A = 3½×A + E Subtract 1¾×A from both sides of the equation above: C + 1¾×A – 1¾×A = 3½×A + E – 1¾×A which becomes eq.4a) C = 1¾×A + E
Hint #7
Substitute 1¼×A for B, 1¾×A for D, and 2¼×A for F in eq.6: A + 1¼×A + C – E = 1¾×A + E + 2¼×A which becomes 2¼×A + C – E = 4×A + E In the above equation, subtract 2¼×A from both sides, and add E to both sides: 2¼×A + C – E – 2¼×A + E = 4×A + E – 2¼×A + E which becomes eq.6a) C = 1¾×A + 2×E
Hint #8
Substitute 1¾×A + E for C (from eq.4a) into eq.6a: 1¾×A + E = 1¾×A + 2×E Subtract 1¾×A and E from each side of the equation above: 1¾×A + E – 1¾×A – E = 1¾×A + 2×E – 1¾×A – E which simplifies to 0 = E
Hint #9
Substitute 0 for E in eq.4a: C = 1¾×A + 0 which makes C = 1¾×A
Solution
Substitute 1¼×A for B, 1¾×A for C and D, 0 for E, and 2¼×A for F in eq.1: A + 1¼×A + 1¾×A + 1¾×A + 0 + 2¼×A = 32 which simplifies to 8×A = 32 Divide both sides of the above equation by 8: 8×A ÷ 8 = 32 ÷ 8 which means A = 4 making B = 1¼×A = 1¼ × 4 = 5 C = D = 1¾×A = 1¾ × 4 = 7 F = 2¼×A = 2¼ × 4 = 9 and ABCDEF = 457709