Puzzle for October 3, 2021 ( )
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.
* DE and BC are 2-digit numbers (not D×E or B×C).
Scratchpad
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Hint #1
eq.4 may be written as: D – E + D = B – D + F which becomes 2×D – E = B – D + F Add E and D to both sides of the equation above: 2×D – E + E + D = B – D + F + E + D which becomes eq.4a) 3×D = B + F + E
Hint #2
eq.5 may be re-written as: C + D = 10×D + E – (10×B + C) which becomes C + D = 10×D + E – 10×B – C In the above equation, subtract D from both sides, and add C to both sides: C + D – D + C = 10×D + E – 10×B – C – D + C which becomes 2×C = 9×D + E – 10×B which may be expressed as eq.5a) 2×C = 3×(3×D) + E – 10×B
Hint #3
In eq.5a, replace 3×D with B + F + E (from eq.4a): 2×C = 3×(B + F + E) + E – 10×B which becomes 2×C = 3×B + 3×F + 3×E + E – 10×B which becomes eq.5b) 2×C = 3×F + 4×E – 7×B
Hint #4
eq.6 may be written as: F = (A + C + E) ÷ 3 Multiply both sides of the equation above by 3: F × 3 = ((A + C + E) ÷ 3) × 3 which becomes 3×F = A + C + E which may be written as eq.6a) 3×F = A + E + C
Hint #5
In eq.6a, replace A + E with B + C (from eq.3): 3×F = B + C + C which becomes eq.6b) 3×F = B + 2×C
Hint #6
In eq.5b, substitute B + 2×C for 3×F (from eq.6b): 2×C = B + 2×C + 4×E – 7×B which becomes 2×C = 2×C + 4×E – 6×B In the equation above, subtract 2×C from each side, and add 6×B to each side: 2×C – 2×C + 6×B = 2×C + 4×E – 6×B – 2×C + 6×B which makes 6×B = 4×E Divide both sides by 4: 6×B ÷ 4 = 4×E ÷ 4 which makes eq.5c) 1½×B = E
Hint #7
Subtract F from both sides of eq.2: E – F = A + F – F which becomes E – F = A Substitute E – F for A in eq.3: B + C = E – F + E which becomes B + C = 2×E – F Multiply both sides of the above equation by 3: 3 × (B + C) = 3 × (2×E – F) which is equivalent to 3×B + 3×C = 3×2×E – 3×F which becomes eq.3a) 3×B + 3×C = 6×E – 3×F
Hint #8
Substitute (1½×B) for E (from eq.5c), and (B + 2×C) for 3×F (from eq.6b) in eq.3a: 3×B + 3×C = 6×(1½×B) – (B + 2×C) which becomes 3×B + 3×C = 9×B – B – 2×C which becomes 3×B + 3×C = 8×B – 2×C In the above equation, subtract 3×B from both sides, and add 2×C to both sides: 3×B + 3×C – 3×B + 2×C = 8×B – 2×C – 3×B + 2×C which makes 5×C = 5×B Divide both sides by 5: 5×C ÷ 5 = 5×B ÷ 5 which makes C = B
Hint #9
Substitute B for C, and 1½×B for E in eq.3: B + B = A + 1½×B which becomes 2×B = A + 1½×B Subtract 1½×B from both sides of the above equation: 2×B – 1½×B = A + 1½×B – 1½×B which makes ½×B = A Multiply both sides by 2: ½×B × 2 = A × 2 which makes B = 2×A and also makes C = B = 2×A
Hint #10
Substitute (2×A) for B in eq.5c: 1½×(2×A) = E which makes 3×A = E
Hint #11
Substitute 3×A for E in eq.2: 3×A = A + F Subtract A from each side of the equation above: 3×A – A = A + F – A which makes 2×A = F
Hint #12
Substitute 2×A for B and F, and 3×A for E in eq.4a: 3×D = 2×A + 2×A + 3×A which makes 3×D = 7×A Divide both sides of the above equation by 3: 3×D ÷ 3 = 7×A ÷ 3 which makes D = 2⅓×A
Solution
Substitute 2×A for B and C and F, 2⅓×A for D, and 3×A for E in eq.1: A + 2×A + 2×A + 2⅓×A + 3×A + 2×A = 37 which simplifies to 12⅓×A = 37 Divide both sides of the above equation by 12⅓: 12⅓×A ÷ 12⅓ = 37 ÷ 12⅓ which means A = 3 making B = C = F = 2×A = 2 × 3 = 6 D = 2⅓×A = 2⅓ × 3 = 7 E = 3×A = 3 × 3 = 9 and ABCDEF = 366796