Puzzle for January 1, 2023 ( )
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
Help Area
Hint #1
In eq.2, replace D with E + F (from eq.4): eq.2a) E + F = A + C
Hint #2
Subtract A from each side of eq.3: C - A = A + B - A which becomes eq.3a) C - A = B
Hint #3
In eq.1, substitute C - A for B (from eq.3a), A + C for D (from eq.2), and A + C for E + F (from eq.2a): A + C - A + C + A + C + A + C = 24 which becomes 2×A + 4×C = 24 Divide both sides of the above equation by 2: (2×A + 4×C) ÷ 2 = 24 ÷ 2 which becomes A + 2×C = 12 Subtract 2×C from both sides: A + 2×C - 2×C = 12 - 2×C which makes eq.1a) A = 12 - 2×C
Hint #4
In eq.6, replace D with A + C (from eq.2), and E + F with A + C (from eq.2a): A × C = A + C + A + C which becomes eq.6a) A × C = 2×A + 2×C
Hint #5
In eq.6a, substitute (12 - 2×C) for A (from eq.1a): (12 - 2×C) × C = 2×(12 - 2×C) + 2×C which becomes 12×C - 2×C² = 24 - 4×C + 2×C which becomes 12×C - 2×C² = 24 - 2×C In the above equation, subtract 12×C from both sides, and add 2×C² to both sides: 12×C - 2×C² - 12×C + 2×C² = 24 - 2×C - 12×C + 2×C² which becomes 0 = 24 - 14×C + 2×C² which may be written as 0 = 2×C² - 14×C + 24 Divide both sides by 2: 0 ÷ 2 = (2×C² - 14×C + 24) ÷ 2 which becomes eq.6b) 0 = C² - 7×C + 12
Hint #6
eq.6b is a quadratic equation in standard form. eq.6b may be solved using the quadratic equation solution formula, or by factoring into the product of two terms. Factoring eq.6b into the product of two terms yields: 0 = (C - 4) × (C - 3) which means C = 3 or C = 4
Hint #7
Begin checking: C = 3 ... Substituting 3 for C in eq.1a would yield: A = 12 - 2×3 which would become A = 12 - 6 which would make A = 6
Hint #8
Finish checking: C = 3 ... Substituting 3 for C, and 6 for A in eq.3a would yield: 3 - 6 = B which would make -3 = B Since B must be non-negative, then: B ≠ -3 which means C ≠ 3 and therefore makes: C = 4
Hint #9
Substitute 4 for C in eq.1a: A = 12 - 2×4 which makes A = 12 - 8 which means A = 4
Hint #10
Substitute 4 for C and A in eq.3a: 4 - 4 = B which makes 0 = B
Hint #11
Substitute 4 for A and C in eq.2: D = 4 + 4 which makes D = 8
Hint #12
Substitute 8 for D in eq.4: 8 = E + F Subtract E from each side of the equation above: 8 - E = E + F - E which becomes eq.4a) 8 - E = F
Hint #13
Substitute 4 for C, (8 - E) for F (from eq.4a), and 8 for D in eq.5: 4 + E - (8 - E) = 8 - E + (8 - E) which becomes 4 + E - 8 + E = 8 - E + 8 - E which becomes -4 + 2×E = 16 - 2×E Add 4 and 2×E to both sides of the above equation: -4 + 2×E + 4 + 2×E = 16 - 2×E + 4 + 2×E which simplifies to 4×E = 20 Divide both sides by 4: 4×E ÷ 4 = 20 ÷ 4 which makes E = 5
Solution
Substitute 5 for E in eq.4a: 8 - 5 = F which makes 3 = F and makes ABCDEF = 404853