Puzzle for May 7, 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
eq.5 may be written as: F – A + B = A + D + E – B In the above equation, replace A + D with B + E (from eq.4): F – A + B = B + E + E – B which becomes F – A + B = 2×E Add A to both sides of the above equation: F – A + B + A = 2×E + A which becomes eq.5a) F + B = 2×E + A
Hint #2
eq.6 may be written as: D + E = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (D + E) = 3 × (B + C + F) ÷ 3 which becomes 3×(D + E) = B + C + F which may be written as eq.6a) 3×D + 3×E = F + B + C
Hint #3
In eq.6a, replace F + B with 2×E + A (from eq.5a): 3×D + 3×E = 2×E + A + C Subtract 2×E from each side of the equation above: 3×D + 3×E – 2×E = 2×E + A + C – 2×E which becomes eq.6b) 3×D + E = A + C
Hint #4
In eq.2, substitute 3×D + E for A + C (from eq.6b): D + F = 3×D + E Subtract D from both sides of the equation above: D + F – D = 3×D + E – D which becomes eq.2a) F = 2×D + E
Hint #5
Substitute 2×D + E for F (from eq.2a) into eq.3: C + D = E + 2×D + E which becomes C + D = 2×E + 2×D Subtract D from both sides of the above equation: C + D – D = 2×E + 2×D – D which becomes eq.3a) C = 2×E + D
Hint #6
Substitute 2×D + E for F (from eq.2a), and 2×E + D for C (from eq.3a) in eq.6a: 3×D + 3×E = 2×D + E + B + 2×E + D which becomes 3×D + 3×E = 3×D + 3×E + B Subtract 3×D and 3×E from both sides of the equation above: 3×D + 3×E – 3×D – 3×E = 3×D + 3×E + B – 3×D – 3×E which makes 0 = B
Hint #7
In eq.5, substitute 2×D + E for F (from eq.2a), and 0 for B: 2×D + E – A + 0 = A – 0 + D + E which becomes 2×D + E – A = A + D + E In the above equation, subtract E and D from both sides, and add A to both sides: 2×D + E – A – E – D + A = A + D + E – E – D + A which simplifies to D = 2×A
Hint #8
Substitute 0 for B, and 2×A for D in eq.4: 0 + E = A + 2×A which makes E = 3×A
Hint #9
Substitute (3×A) for E, and 2×A for D in eq.3a: C = 2×(3×A) + 2×A which becomes C = 6×A + 2×A which makes C = 8×A
Hint #10
Substitute (2×A) for D, and 3×A for E in eq.2a: F = 2×(2×A) + 3×A which becomes F = 4×A + 3×A which makes F = 7×A
Solution
Substitute 0 for B, 8×A for C, 2×A for D, 3×A for E, and 7×A for F in eq.1: A + 0 + 8×A + 2×A + 3×A + 7×A = 21 which simplifies to 21×A = 21 Divide both sides of the above equation by 21: 21×A ÷ 21 = 21 ÷ 21 which means A = 1 making C = 8×A = 8 × 1 = 8 D = 2×A = 2 × 1 = 2 E = 3×A = 3 × 1 = 3 F = 7×A = 7 × 1 = 7 and ABCDEF = 108237