Puzzle for May 8, 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
Add E and D to both sides of eq.2: D – E + E + D = A – D + E + D which becomes eq.2a) 2×D = A + E Add D and E to both sides of eq.2: C – D + E + D + E = A + D – E + D + E which becomes eq.4a) C + 2×E = A + 2×D
Hint #2
In eq.4a, replace 2×D with A + E (from eq.2a): C + 2×E = A + A + E which becomes C + 2×E = 2×A + E Subtract E from each side of the above equation: C + 2×E – E = 2×A + E – E which becomes eq.4b) C + E = 2×A
Hint #3
In the eq.3, replace C + E with 2×A (from eq.4b): A + B = 2×A + F – A which becomes A + B = A + F Subtract A from each side of the equation above: A + B – A = A + F – A which makes B = F
Hint #4
In eq.5, substitute B for F: D – E + B = A – B + E Add E and B to both sides of the above equation: D – E + B + E + B = A – B + E + E + B which becomes eq.5a) D + 2×B = A + 2×E
Hint #5
eq.6 may be written as: A – F = (B + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (A – F) = 4 × (B + C + D + F) ÷ 4 which becomes eq.6a) 4×A – 4×F = B + C + D + F
Hint #6
Substitute B for F in eq.6a: 4×A – 4×B = B + C + D + B which becomes 4×A – 4×B = 2×B + C + D which may be written as eq.6b) 4×A – 4×B = D + 2×B + C
Hint #7
Substitute A + 2×E for D + 2×B (from eq.5a) into eq.6b: 4×A – 4×B = A + 2×E + C Subtract A from both sides of the above equation: 4×A – 4×B – A = A + 2×E + C – A which becomes 3×A – 4×B = 2×E + C which may be written as eq.6c) 3×A – 4×B = C + 2×E
Hint #8
Substitute A + 2×D for C + 2×E (from eq.4a) in eq.6c: 3×A – 4×B = A + 2×D In the above equation, subtract A from both sides, and add 4×B to both sides: 3×A – 4×B – A + 4×B = A + 2×D – A + 4×B which becomes 2×A = 2×D + 4×B Divide both sides by 2: 2×A ÷ 2 = (2×D + 4×B) ÷ 2 which becomes eq.6d) A = D + 2×B
Hint #9
Substitute A for D + 2×B (from eq.6d) in eq.5a: A = A + 2×E Subtract A from each side of the equation above: A – A = A + 2×E – A which makes 0 = 2×E which means 0 = E
Hint #10
Substitute 0 for E in eq.2a: 2×D = A + 0 which makes eq.2b) 2×D = A
Hint #11
Substitute 2×D for A in eq.6d: 2×D = D + 2×B Subtract D from each side of the above equation: 2×D – D = D + 2×B – D which makes D = 2×B
Hint #12
Substitute (2×B) for D in eq.2b: 2×(2×B) = A which makes 4×B = A
Hint #13
Substitute 0 for E, and 4×B for A in eq.4b: C + 0 = 2×(4×B) which makes C = 8×B
Solution
Substitute 4×B for A, 8×B for C, 2×B for D, 0 for E, and B for F in eq.1: 4×B + B + 8×B + 2×B + 0 + B = 16 which simplifies to 16×B = 16 Divide both sides of the above equation by 16: 16×B ÷ 16 = 16 ÷ 16 which means B = 1 making A = 4×B = 4 × 1 = 4 C = 8×B = 8 × 1 = 8 D = 2×B = 2 × 1 = 2 F = B = 1 and ABCDEF = 418201