Puzzle for March 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
eq.5 may be written as: F = (A + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + C + E) ÷ 3 which becomes eq.5a) 3×F = A + C + E
Hint #2
Add E to both sides of eq.4: A + B + C – F + E = E + F – B + E which becomes A + B + C – F + E = 2×E + F – B which may be written as A + C + E + B – F = 2×E + F – B In the above equation, replace A + C + E with 3×F (from eq.5a): 3×F + B – F = 2×E + F – B which becomes 2×F + B = 2×E + F – B Subtract F from both sides, and add B to both sides: 2×F + B – F + B = 2×E + F – B – F + B which becomes eq.4a) F + 2×B = 2×E
Hint #3
eq.6 may be written as: E = (B + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × (B + C + D) ÷ 3 which becomes eq.6a) 3×E = B + C + D
Hint #4
In eq.6a, substitute E + F for C + D (from eq.2): 3×E = B + E + F Subtract E from each side of the above equation: 3×E – E = B + E + F – E which becomes eq.6b) 2×E = B + F
Hint #5
Substitute B + F for 2×E (from eq.6b) into eq.4a: F + 2×B = B + F Subtract B and F from both sides of the equation above: F + 2×B – B – F = B + F – B – F which makes B = 0
Hint #6
Substitute 0 for B in eq.6b: 2×E = 0 + F which makes eq.4b) 2×E = F
Hint #7
Substitute (2×E) for F (from eq.4b) in eq.5a: 3×(2×E) = A + C + E which becomes 6×E = A + C + E Subtract A and E from each side of the equation above: 6×E – A – E = A + C + E – A – E which becomes eq.5b) 5×E – A = C
Hint #8
Substitute 0 for B, and 5×E – A for C (from eq.5b) in eq.3: A + D = 0 + 5×E – A which becomes A + D = 5×E – A Add A to both sides of the above equation: A + D + A = 5×E – A + A which becomes eq.3a) 2×A + D = 5×E
Hint #9
Substitute 0 for B, and 5×E – A for C (from eq.5b) in eq.6a: 3×E = 0 + 5×E – A + D which becomes 3×E = 5×E – A + D In the above equation, subtract 3×E from both sides, and add A to both sides: 3×E – 3×E + A = 5×E – A + D – 3×E + A which becomes eq.6c) A = 2×E + D
Hint #10
Substitute (2×E + D) for A (from eq.6c) into eq.3a: 2×(2×E + D) + D = 5×E which becomes 4×E + 2×D + D = 5×E which becomes 4×E + 3×D = 5×E Subtract 4×E from both sides of the equation above: 4×E + 3×D – 4×E = 5×E – 4×E which makes 3×D = E
Hint #11
Substitute (3×D) for E in eq.4b: 2×(3×D) = F which makes 6×D = F
Hint #12
Substitute (3×D) for E in eq.6c: A = 2×(3×D) + D which becomes A = 6×D + D which makes A = 7×D
Hint #13
Substitute (3×D) for E, and 7×D for A in eq.5b: 5×(3×D) – 7×D = C which becomes 15×D – 7×D = C which makes 8×D = C
Solution
Substitute 7×D for A, 0 for B, 8×D for C, 3×D for E, and 6×D for F in eq.1: 7×D + 0 + 8×D + D + 3×D + 6×D = 25 which simplifies to 25×D = 25 Divide both sides of the above equation by 25: 25×D ÷ 25 = 25 ÷ 25 which means D = 1 making A = 7×D = 7 × 1 = 7 C = 8×D = 8 × 1 = 8 E = 3×D = 3 × 1 = 3 F = 6×D = 6 × 1 = 6 and ABCDEF = 708136