Puzzle for August 29, 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
eq.6 may be written as: D = (A + C + E) ÷ 3 Multiply both sides of the equation above by 3: 3 × D = 3 × (A + C + E) ÷ 3 which becomes 3×D = A + C + E Add A to both sides of the above equation: 3×D + A = A + C + E + A which becomes eq.6a) 3×D + A = 2×A + C + E
Hint #2
Add A to both sides of eq.4: E + F – A + A = A + C + A which becomes eq.4a) E + F = 2×A + C In eq.6a, replace 2×A + C with E + F (from eq.4a): 3×D + A = E + F + E which becomes eq.6b) 3×D + A = 2×E + F
Hint #3
Add E and F to both sides of eq.3: D – E + E + F = E – F + E + F which becomes eq.3a) D + F = 2×E In eq.6b, substitute D + F for 2×E (from eq.3a): 3×D + A = D + F + F which becomes 3×D + A = D + 2×F Subtract 3×D from both sides of the equation above: 3×D + A – 3×D = D + 2×F – 3×D which becomes eq.6c) A = 2×F – 2×D
Hint #4
Subtract B from both sides of eq.2: B + C – B = F – B which becomes C = F – B Substitute F – B for C in eq.5: F – B – D + F = A + D which becomes 2×F – B – D = A + D In the above equation, subtract D from both sides, add B to both sides: 2×F – B – D – D + B = A + D – D + B which becomes eq.5a) 2×F – 2×D = A + B
Hint #5
Substitute A for 2×F – 2×D (from eq.6c) in eq.5a: A = A + B Subtract A from each side of the equation above: A – A = A + B – A which makes 0 = B
Hint #6
Substitute 0 for B in eq.2: 0 + C = F which makes C = F
Hint #7
Substitute C for F in eq.4a: E + C = 2×A + C Subtract C from both sides of the above equation: E + C – C = 2×A + C – C which makes E = 2×A
Hint #8
Substitute (2×A) for E in eq.3a: D + F = 2×(2×A) which becomes D + F = 4×A Subtract D from each side of the above equation: D + F – D = 4×A – D which becomes eq.3b) F = 4×A – D
Hint #9
Substitute (4×A – D) for F (from eq.3b) in eq.6c: A = 2×(4×A – D) – 2×D which becomes A = 8×A – 2×D – 2×D which becomes A = 8×A – 4×D In the equation above, add 4×D to both sides, and subtract A from both sides: A + 4×D – A = 8×A – 4×D + 4×D – A which becomes 4×D = 7×A Divde both sides by 4: 4×D ÷ 4 = 7×A ÷ 4 which makes D = 1¾×A
Hint #10
Substitute 1¾×A for D in eq.3b: F = 4×A – 1¾×A which makes F = 2¼×A and also makes C = F = 2¼×A
Solution
Substitute 0 for B, 2¼×A for C and F, 1¾×A for D, and 2×A for E in eq.1: A + 0 + 2¼×A + 1¾×A + 2×A + 2¼×A = 37 which simplifies to 9¼×A = 37 Divide both sides of the equation above by 9¼: 9¼×A ÷ 9¼ = 37 ÷ 9¼ which means A = 4 making C = F = 2¼×A = 2¼ × 4 = 9 D = 1¾×A = 1¾ × 4 = 7 E = 2×A = 2 × 4 = 8 and ABCDEF = 409789