Puzzle for March 12, 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: A = (C + F) ÷ 2 Multiply both sides of the equation above by 2: 2 × A = 2 × (C + F) ÷ 2 which becomes eq.6a) 2×A = C + F
Hint #2
Add A and C to both sides of eq.5: F – A + A + C = A + B – C + A + C which becomes F + C = 2×A + B In the above equation, replace 2×A with C + F (from eq.6a): F + C = C + F + B Subtract C and F from both sides: F + C – C – F = C + F + B – C – F which simplifies to 0 = B
Hint #3
Subtract C and A from both sides of eq.2: C + D – C – A = A + B + F – C – A which becomes D – A = B + F – C In the equation above, replace B with 0: D – A = 0 + F – C which becomes eq.2a) D – A = F – C
Hint #4
In eq.3, replace B with 0, and F – C with D – A (from eq.2a): E – A – 0 = D – A which becomes E – A = D – A Add A to both sides of the above equation: E – A + A = D – A + A which makes E = D
Hint #5
Add A and D to both sides of eq.4: D + E – A + A + D = A + C – D + F + A + D which becomes 2×D + E = 2×A + C + F Substitute D for E in the equation above: 2×D + D = 2×A + C + F which becomes eq.4a) 3×D = 2×A + C + F
Hint #6
In eq.4a, substitute 2×A for C + F (from eq.6a): 3×D = 2×A + 2×A which becomes 3×D = 4×A Divide both sides of the above equation by 4: 3×D ÷ 4 = 4×A ÷ 4 which makes ¾×D = A
Hint #7
Substitute ¾×D for A, and 0 for B in eq.2: C + D = ¾×D + 0 + F which becomes C + D = ¾×D + F Subtract ¾×D from each side of the above equation: C + D – ¾×D = ¾×D + F – ¾×D which becomes eq.2b) C + ¼×D = F
Hint #8
Substitute (¾×D) for A, and C + ¼×D for F (from eq.2b) in eq.6a: 2×(¾×D) = C + C + ¼×D which becomes 1½×D = 2×C + ¼×D Subtract ¼×D from each side of the above equation: 1½×D – ¼×D = 2×C + ¼×D – ¼×D which makes 1¼×D = 2×C Divide both sides by 2: 1¼×D ÷ 2 = 2×C ÷ 2 which makes ⅝×D = C
Hint #9
Substitute ⅝×D for C in eq.2b: ⅝×D + ¼×D = F which makes ⅞×D = F
Solution
Substitute ¾×D for A, 0 for B, ⅝×D for C, D for E, and ⅞×D for F in eq.1: ¾×D + 0 + ⅝×D + D + D + ⅞×D = 34 which simplifies to 4¼×D = 34 Divide both sides of the above equation by 4¼: 4¼×D ÷ 4¼ = 34 ÷ 4¼ which means D = 8 making A = ¾×D = ¾ × 8 = 6 C = ⅝×D = ⅝ × 8 = 5 E = D = 8 F = ⅞×D = ⅞ × 8 = 7 and ABCDEF = 605887