Puzzle for June 26, 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
Subtract the left and right sides of eq.4 from the left and right sides of eq.5, respectively: B + C – D – (A + B – D) = A – C + D – F – (C + D – F) which becomes B + C – D – A – B + D = A – C + D – F – C – D + F which becomes C – A = A – 2×C Add A and 2×C to both sides of the above equation: C – A + A + 2×C = A – 2×C + A + 2×C which makes 3×C = 2×A Divide both sides by 2: 3×C ÷ 2 = 2×A ÷ 2 which makes A = 1½×C
Hint #2
Multiply both sides of eq.4 by (–1): –1 × (A + B – D) = –1 × (C + D – F) which becomes –A – B + D = –C – D + F Add B and D to both sides of the above equation: –A – B + D + B + D = –C – D + F + B + D which becomes –A + 2×D = –C + F + B which is the same as eq.4a) 2×D – A = B – C + F
Hint #3
In eq.4a, replace A with 1½×C, and replace B – C + F with C – D (from eq.2): 2×D – 1½×C = C – D Add 1½×C and D to both sides of the above equation: 2×D – 1½×C + 1½×C + D = C – D + 1½×C + D which becomes 3×D = 2½×C Divide both sides by 3: 3×D ÷ 3 = 2½×C ÷ 3 which makes D = ⅚×C
Hint #4
eq.6 may be written as: (B + C + E + F) ÷ 4 = (A + C + F) ÷ 3 Multiply both sides of the above equation by 12: 12 × (B + C + E + F) ÷ 4 = 12 × (A + C + F) ÷ 3 which becomes 3 × (B + C + E + F) = 4 × (A + C + F) which is equivalent to 3×B + 3×C + 3×E + 3×F = 4×A + 4×C + 4×F Subtract 3×C and 3×F from both sides: 3×B + 3×C + 3×E + 3×F – 3×C – 3×F = 4×A + 4×C + 4×F – 3×C – 3×F which becomes 3×B + 3×E = 4×A + C + F which may be written as eq.6a) 3×(B + E) = 4×A + C + F
Hint #5
Substitute 1½×C for A, and ⅚×C for D in eq.3: 1½×C + ⅚×C = B + E + F which becomes 2⅓×C = B + E + F Subtract F from each side of the above equation: 2⅓×C – F = B + E + F – F which becomes eq.3a) 2⅓×C – F = B + E
Hint #6
Substitute 2⅓×C – F for B + E (from eq.3a), and (1½×C) for A in eq.6a: 3×(2⅓×C – F) = 4×(1½×C) + C + F which becomes 7×C – 3×F = 6×C + C + F which becomes 7×C – 3×F = 7×C + F In the above equation, subtract 7×C from both sides, and add 3×F to both sides: 7×C – 3×F – 7×C + 3×F = 7×C + F – 7×C + 3×F which becomes 0 = 4×F which means 0 = F
Hint #7
Substitute (⅚×C) for D, 1½×C for A, and 0 for F in eq.4a: 2×(⅚×C) – 1½×C = B – C + 0 which becomes 1⅔×C – 1½×C = B – C which becomes ⅙×C = B – C Add C to both sides of the equation above: ⅙×C + C = B – C + C which makes 1⅙×C = B
Hint #8
Substitute 0 for F, and 1⅙×C for B in eq.3a: 2⅓×C – 0 = 1⅙×C + E which becomes 2⅓×C = 1⅙×C + E Subtract 1⅙×C fom each side of the above equation: 2⅓×C – 1⅙×C = 1⅙×C + E – 1⅙×C which makes 1⅙×C = E
Solution
Substitute 1½×C for A, 1⅙×C for B and E, ⅚×C for D, and 0 for F in eq.1: 1½×C + 1⅙×C + C + ⅚×C + 1⅙×C + 0 = 34 which simplifies to 5⅔×C = 34 Divide both sides of the above equation by 5⅔: 5⅔×C ÷ 5⅔ = 34 ÷ 5⅔ which means C = 6 making A = 1½×C = 1½ × 6 = 9 B = E = 1⅙×C = 1⅙ × 6 = 7 D = ⅚×C = ⅚ × 6 = 5 and ABCDEF = 976570