Puzzle for June 11, 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..
Scratchpad
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Hint #1
Add A to both sides of eq.4: A + F + A = C + E – A + A which becomes 2×A + F = C + E In eq.5, replace C + E with 2×A + F: 2×A + F = A + B – E + F In the above equation, subtract F and A from both sides, and add E to both sides: 2×A + F – F – A + E = A + B – E + F – F – A + E which becomes eq.5a) A + E = B
Hint #2
Add E and C to each side of eq.3: C – E + E + C = A – C + F + E + C which becomes 2×C = A + F + E which is the same as 2×C = A + E + F In the above equation, replace A + E with B (from eq.5a): eq.3a) 2×C = B + F
Hint #3
In eq.3a, substitute C + F for B (from eq.2): 2×C = C + F + F which becomes 2×C = C + 2×F Subtract C from both sides of the equation above: 2×C – C = C + 2×F – C which makes C = 2×F
Hint #4
Substitute 2×F for C in eq.2: B = 2×F + F which makes B = 3×F
Hint #5
Substitute 2×F for C, and 3×F for B in eq.5: 2×F + E = A + 3×F – E + F which becomes 2×F + E = A + 4×F – E In the above equation, subtract 4×F from both sides, and add E to both sides: 2×F + E – 4×F + E = A + 4×F – E – 4×F + E which becomes eq.5b) 2×E – 2×F = A
Hint #6
Substitute 2×F for C in eq.4: A + F = 2×F + E – A In the above equation, subtract F from both sides, and add A to both sides: A + F – F + A = 2×F + E – A – F + A which becomes eq.4a) 2×A = F + E
Hint #7
Substitute (2×E – 2×F) for A (from eq.5b) in eq.4a: 2×(2×E – 2×F) = F + E which becomes 4×E – 4×F = F + E In the above equation, add 4×F to both sides, and subtract E from both sides: 4×E – 4×F + 4×F – E = F + E + 4×F – E which becomes 3×E = 5×F Divide both sides by 3: 3×E ÷ 3 = 5×F ÷ 3 which makes E = 1⅔×F
Hint #8
Substitute (1⅔×F) for E in eq.5b: 2×(1⅔×F) – 2×F = A which becomes 3⅓×F – 2×F = A which makes 1⅓×F = A
Hint #9
eq.6 may be written as: E = (B + C + D) ÷ 3 Multiply both sides of the above equation by 3: E × 3 = ((B + C + D) ÷ 3) × 3 which becomes eq.6a) 3×E = B + C + D
Hint #10
Substitute (1⅔×F) for E, 3×F for B, and 2×F for C in eq.6a: 3×(1⅔×F) = 3×F + 2×F + D which becomes 5×F = 5×F + D Subtract 5×F from both sides of the above equation: 5×F – 5×F = 5×F + D – 5×F which makes 0 = D
Solution
Substitute 1⅓×F for A, 3×F for B, 2×F for C, 0 for D, and 1⅔×F for E in eq.1: 1⅓×F + 3×F + 2×F + 0 + 1⅔×F + F = 27 which simplifies to 9×F = 27 Divide both sides of the above equation by 9: 9×F ÷ 9 = 27 ÷ 9 which means F = 3 making A = 1⅓×F = 1⅓ × 3 = 4 B = 3×F = 3 × 3 = 9 C = 2×F = 2 × 3 = 6 E = 1⅔×F = 1⅔ × 3 = 5 and ABCDEF = 496053