Puzzle for July 21, 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
In eq.4, replace C with A + B (from eq.2): B + A + B = A + E which becomes A + 2×B = A + E Subtract A from each side of the above equation: A + 2×B – A = A + E – A which makes 2×B = E
Hint #2
In eq.6, replace A + B with C (from eq.2): D + F = C + C which becomes eq.6a) D + F = 2×C
Hint #3
In eq.6a, substitute (D + E) for C (from eq.3): D + F = 2×(D + E) which becomes D + F = 2×D + 2×E Subtract D from both sides of the above equation: D + F – D = 2×D + 2×E – D which becomes eq.6b) F = D + 2×E
Hint #4
Substitute D + 2×E for F (from eq.6b) in eq.5: E + D + 2×E = A + D which becomes D + 3×E = A + D Subtract D from each side of the above equation: D + 3×E – D = A + D – D which makes eq.5a) 3×E = A
Hint #5
Substitute (2×B) for E in in eq.5a: 3×(2×B) = A which makes 6×B = A
Hint #6
Substitute 6×B for A in eq.2: C = 6×B + B which makes C = 7×B
Hint #7
Substitute 2×B for E, and 7×B for C in eq.3: D + 2×B = 7×B Subtract 2×B from each side of the equation above: D + 2×B – 2×B = 7×B – 2×B which makes D = 5×B
Hint #8
Substitute 5×B for D, and (2×B) for E in eq.6b: F = 5×B + 2×(2×B) which becomes F = 5×B + 4×B which makes F = 9×B
Solution
Substitute 6×B for A, 7×B for C, 5×B for D, 2×B for E, and 9×B for F in eq.1: 6×B + B + 7×B + 5×B + 2×B + 9×B = 30 which simplifies to 30×B = 30 Divide both sides of the above equation by 30: 30×B ÷ 30 = 30 ÷ 30 which means B = 1 making A = 6×B = 6 × 1 = 6 C = 7×B = 7 × 1 = 7 D = 5×B = 5 × 1 = 5 E = 2×B = 2 × 1 = 2 F = 9×B = 9 × 1 = 9 and ABCDEF = 617529