Puzzle for October 28, 2019 ( )
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, substitute (B + E) for D (from eq.2): E – B = C – (B + E) which is equivalent to E – B = C – B – E Add B and E to both sides of the above equation: E – B + B + E = C – B – E + B + E which becomes eq.4a) 2×E = C
Hint #2
In eq.6, replace C with 2×E: 2×E – B = B + E In the equation above, add B to both sides, and subtract E from both sides: 2×E – B + B – E = B – E + B – E which makes E = 2×B
Hint #3
In eq.4a, replace E with (2×B): 2×(2×B) = C which makes 4×B = C
Hint #4
In eq.2, replace E with 2×B: B + 2×B = D which makes 3×B = D
Hint #5
In eq.3, substitute 4×B for C: 4×B – F = B In the equation above, add F to both sides, and subtract B from each side: 4×B – F + F – B = B + F – B which makes 3×B = F
Hint #6
Substitute 2×B for E, 4×B for C, and 3×B for F in eq.5: A + 2×B = 4×B + 3×B – A which becomes A + 2×B = 7×B – A In the above equation, add A to both sides, and subtract 2×B from each side: A + 2×B + A – 2×B = 7×B – A + A – 2×B which means 2×A = 5×B Divide both sides by 2: 2×A ÷ 2 = 5×B ÷ 2 which makes A = 2½×B
Solution
Substitute 2½×B for A, 4×B for C, 3×B for D and F, and 2×B for E in eq.1: 2½×B + B + 4×B + 3×B + 2×B + 3×B = 31 which simplifies to 15½×B = 31 Divide both sides of the equation above by 15½: 15½×B ÷ 15½×B = 31 ÷ 15½ which means B = 2 making A = 2½×B = 2½ × 2 = 5 C = 4×B = 4 × 2 = 8 D = F = 3×B = 3 × 2 = 6 E = 2×B = 2 × 2 = 4 and ABCDEF = 528646