Puzzle for February 6, 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 F and E to both sides of eq.2: E – F + F + E = B – D – E + F + E which becomes 2×E = B – D + F which may be written as eq.2a) 2×E = B + F – D
Hint #2
In eq.2a, replace B + F with A + D – F (from eq.4): 2×E = A + D – F – D which becomes 2×E = A – F Add F to both sides of the above equation: 2×E + F = A – F + F which becomes eq.2b) 2×E + F = A
Hint #3
Add A and E to both sides of eq.2a: 2×E + A + E = B + F – D + A + E which is equivalent to 3×E + A = F – D + A + B + E In the above equation, replace A + B + E with C + D (from eq.3): 3×E + A = F – D + C + D which becomes 3×E + A = F + C which is the same as eq.2c) 3×E + A = C + F
Hint #4
In eq.2c, substitute 2×E + F for A (from eq.2b): 3×E + 2×E + F = C + F which becomes 5×E + F = C + F Subtract F from each side of the equation above: 5×E + F – F = C + F – F which makes 5×E = C
Hint #5
eq.6 may be written as: E + F = (A + B + C) ÷ 3 Multiply both sides of the above equation by 3: 3 × (E + F) = 3 × (A + B + C) ÷ 3 which becomes eq.6a) 3×E + 3×F = A + B + C
Hint #6
Add F to both sides of eq.6a: 3×E + 3×F + F = A + B + C + F which becomes 3×E + 4×F = A + B + C + F Substitute 3×E + A for C + F (from eq.2c) into the equation above: 3×E + 4×F = A + B + 3×E + A which becomes 3×E + 4×F = 2×A + B + 3×E Subtract 3×E and 2×A from each side of the above equation: 3×E + 4×F – 3×E – 2×A = 2×A + B + 3×E – 3×E – 2×A which simplifies to eq.6b) 4×F – 2×A = B
Hint #7
Substitute (2×E + F) for A (from eq.2b) in eq.6b: 4×F – 2×(2×E + F) = B which becomes 4×F – 4×E – 2×F = B which becomes eq.6c) 2×F – 4×E = B
Hint #8
Substitute 2×F – 4×E for B (from eq.6c) in eq.2: E – F = 2×F – 4×E – D – E which becomes E – F = 2×F – 5×E – D In the above equation, subtract E from both sides, and add F and D to both sides: E – F – E + F + D = 2×F – 5×E – D – E + F + D which becomes eq.2d) D = 3×F – 6×E
Hint #9
In eq.5, substitute (3×F – 6×E) for D (from eq.2d), 2×F – 4×E for B (from eq.6c), and 5×E for C: (3×F – 6×E) – E = 2×F – 4×E + 5×E – (3×F – 6×E) + E + F which becomes 3×F – 7×E = 2×F + E – 3×F + 6×E + E + F which becomes 3×F – 7×E = 8×E Add 7×E to both sides of the above equation: 3×F – 7×E + 7×E = 8×E + 7×E which makes 3×F = 15×E Divide both sides by 3: 3×F ÷ 3 = 15×E ÷ 3 which makes F = 5×E
Hint #10
Substitute (5×E) for F in eq.6c: 2×(5×E) – 4×E = B which becomes 10×E – 4×E = B which makes 6×E = B
Hint #11
Substitute (5×E) for F in eq.2d: D = 3×(5×E) – 6×E which becomes D = 15×E – 6×E which makes D = 9×E
Hint #12
Substitute 5×E for F in eq.2b: 2×E + 5×E = A which makes 7×E = A
Solution
Substitute 7×E for A, 6×E for B, 5×E for C and F, and 9×E for D in eq.1: 7×E + 6×E + 5×E + 9×E + E + 5×E = 33 which simplifies to 33×E = 33 Divide both sides of the above equation by 33: 33×E ÷ 33 = 33 ÷ 33 which means E = 1 making A = 7×E = 7 × 1 = 7 B = 6×E = 6 × 1 = 6 C = F = 5×E = 5 × 1 = 5 D = 9×E = 9 × 1 = 9 and ABCDEF = 765915