Puzzle for July 9, 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.6, respectively: B + C – D – (C – B) = E + F – C – (F – C – D) which becomes B + C – D – C + B = E + F – C – F + C + D which becomes 2×B – D = E + D Subtract D from each side of the equation above: 2×B – D – D = E + D – D which becomes eq.6a) 2×B – 2×D = E which is equivalent to eq.6b) 2×(B – D) = E
Hint #2
In eq.2, replace E with 2×B – 2×D (from eq.6a): D + 2×B – 2×D = A + B which becomes 2×B – D = A + B Subtract B from each side of the above equation: 2×B – D – B = A + B – B which becomes eq.2a) B – D = A
Hint #3
In eq.6b, replace B – D with A (from eq.2a): 2×(A) = E which is the same as eq.6c) 2×A = E
Hint #4
In eq.3, substitute 2×A for E (from eq.6c): F – C = 2×A – F Add C and F to both sides of the above equation: F – C + C + F = 2×A – F + C + F which becomes eq.3a) 2×F = 2×A + C
Hint #5
In eq.4, add B and C to both sides, and subtract F from both sides: C – B + B + C – F = F – C – D + B + C – F which becomes 2×C – F = –D + B which is the same as 2×C – F = B – D Substitute A for B – D (from eq.2a) in the above equation: eq.4a) 2×C – F = A
Hint #6
Substitute (2×C – F) for A (from eq.4a) into eq.3a: 2×F = 2×(2×C – F) + C which becomes 2×F = 4×C – 2×F + C which becomes 2×F = 5×C – 2×F Add 2×F to both sides of the above equation: 2×F + 2×F = 5×C – 2×F + 2×F which becomes 4×F = 5×C Divide both sides by 4: 4×F ÷ 4 = 5×C ÷ 4 which makes F = 1¼×C
Hint #7
Substitute 1¼×C for F in eq.4a: 2×C – 1¼×C = A which makes ¾×C = A
Hint #8
Substitute (¾×C) for A in eq.6c: 2×(¾×C) = E which makes 1½×C = E
Hint #9
Substitute ¾×C for A, 1½×C for E, and 1¼×C for F in eq.5: ¾×C + 1½×C = C + D + 1¼×C which becomes 2¼×C = 2¼×C + D Subtract 2¼×C from each side of the equation above: 2¼×C – 2¼×C = 2¼×C + D – 2¼×C which makes 0 = D
Hint #10
Substitute 0 for D, and ¾×C for A in eq.2a: B – 0 = ¾×C which makes B = ¾×C
Solution
Substitute ¾×C for A and B, 0 for D, 1½×C for E, and 1¼×C for F in eq.1: ¾×C + ¾×C + C + 0 + 1½×C + 1¼×C = 21 which simplifies to 5¼×C = 21 Divide both sides of the above equation by 5¼: 5¼×C ÷ 5¼ = 21 ÷ 5¼ which means C = 4 making A = B = ¾×C = ¾ × 4 = 3 E = 1½×C = 1½ × 4 = 6 F = 1¼×C = 1¼ × 4 = 5 and ABCDEF = 334065