Puzzle for June 29, 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 B + E (from eq.3), and F with D + E (from eq.2): B + E + D = A + D + E Subtract D and E from each side of the equation above: B + E + D – D – E = A + D + E – D – E which simplifies to B = A
Hint #2
In eq.5, replace A with B, C with B + E (from eq.3), and F with D + E (from eq.2): B + B + D = B + E + D + E which becomes 2×B + D = B + 2×E + D Subtract B and D from each side of the equation above: 2×B + D – B – D = B + 2×E + D – B – D which makes B = 2×E and also makes A = B = 2×E
Hint #3
In eq.3, substitute 2×E for B: 2×E + E = C which makes 3×E = C
Hint #4
eq.6 may be written as: F = (A + B + C) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + B + C) ÷ 3 which becomes eq.6a) 3×F = A + B + C
Hint #5
Substitute 2×E for A and B, and 3×E for C in eq.6a: 3×F = 2×E + 2×E + 3×E which becomes 3×F = 7×E Divide both sides of the above equation by 3: 3×F ÷ 3 = 7×E ÷ 3 which makes F = 2⅓×E
Hint #6
Substitute 2⅓×E for F in eq.2: D + E = 2⅓×E Subtract E from each side of the equation above: D + E – E = 2⅓×E – E which makes D = 1⅓×E
Solution
Substitute 2×E for A and B, 3×E for C, 1⅓×E for D, and 2⅓×E for F in eq.1: 2×E + 2×E + 3×E + 1⅓×E + E + 2⅓×E = 35 which simplifies to 11⅔×E = 35 Divide both sides of the above equation by 11⅔: 11⅔×E ÷ 11⅔ = 35 ÷ 11⅔ which means E = 3 making A = B = 2×E = 2 × 3 = 6 C = 3×E = 3 × 3 = 9 D = 1⅓×E = 1⅓ × 3 = 4 F = 2⅓×E = 2⅓ × 3 = 7 and ABCDEF = 669437