Puzzle for February 23, 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.
* BC is a 2-digit number (not B×C).
Scratchpad
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Hint #1
Add F to both sides of eq.2: B + F = A – F + F which becomes eq.2a) B + F = A In eq.3, replace A with B + F (from eq.2a): C = B + F + F which becomes eq.3a) C = B + 2×F
Hint #2
Add F to both sides of eq.4: D – F + F = B + F + F which becomes eq.4a) D = B + 2×F In eq.3a, replace B + 2×F with D (from eq.4a): C = D
Hint #3
eq.6 may be written as: A = (B + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × A = 4 × (B + C + D + F) ÷ 4 which becomes eq.6a) 4×A = B + C + D + F
Hint #4
In eq.6a, substitute (B + F) for A (from eq.2a), and B + 2×F for C (from eq.3a) and for D (from eq.4a): 4×(B + F) = B + B + 2×F + B + 2×F + F which becomes 4×B + 4×F = 3×B + 5×F Subtract 4×F and 3×B from both sides of the above equation: 4×B + 4×F – 4×F – 3×B = 3×B + 5×F – 4×F – 3×B which simplifies to B = F
Hint #5
Substitute B for F in eq.3a: C = B + 2×B which makes C = 3×B and also makes D = C = 3×B
Hint #6
Substitute B for F in eq.2a: A = B + B which makes A = 2×B
Hint #7
eq.5 may be written as: 10×B + C – E = D + E + F Substitute 3×B for C and D, and B for F in the above equation: 10×B + 3×B – E = 3×B + E + B which becomes 13×B – E = 4×B + E Add E to both sides, and subtract 4×B from both sides: 13×B – E + E – 4×B = 4×B + E + E – 4×B which becomes 9×B = 2×E Divide both sides of the above equation by 2: 9×B ÷ 2 = 2×E ÷ 2 which makes 4½×B = E
Solution
Substitute 2×B for A, 3×B for C and D, 4½×B for E, and B for F in eq.1: 2×B + B + 3×B + 3×B + 4½×B + B = 29 which simplifies to 14½×B = 29 Divide both sides of the above equation by 14½: 14½×B ÷ 14½ = 29 ÷ 14½ which means B = 2 making A = 2×B = 2 × 2 = 4 C = D = 3×B = 3 × 2 = 6 E = 4½×B = 4½ × 2 = 9 F = B = 2 and ABCDEF = 426692