Puzzle for June 17, 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.
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Hint #1
Add A and F to both sides of eq.4: F - A + A + F = D - F + A + F which becomes eq.4a) 2×F = D + A
Hint #2
Subtract the left and right sides of eq.2 from the left and right sides of eq.4a, respectively: 2×F - (E + F) = D + A - (A + B) which becomes 2×F - E - F = D + A - A - B which becomes F - E = D - B Add E and B to both sides of the above equation: F - E + E + B = D - B + E + B which becomes eq.4b) F + B = D + E
Hint #3
Add C to both sides of eq.4b: F + B + C = D + E + C which may be written as eq.4c) B + C + F = D + E + C
Hint #4
In eq.4c, replace B + C with D + F (from eq.3): D + F + F = D + E + C which becomes D + 2×F = D + E + C Subtract D from each side of the equation above: D + 2×F - D = D + E + C - D which becomes 2×F = E + C which may be written as eq.4d) 2×F = C + E
Hint #5
In eq.5, replace C + E with 2×F (from eq.4d): F - C = 2×F - F which becomes F - C = F Subtract F from each side of the above equation: F - C - F = F - F which makes -C = 0 which means C = 0
Hint #6
In eq.4d, substitute 0 for C: 2×F = 0 + E which makes 2×F = E
Hint #7
Substitute 2×F for E in eq.2: 2×F + F = A + B which becomes eq.2a) 3×F = A + B
Hint #8
eq.6 may be written as: D = (A + B + C + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × D = 4 × (A + B + C + E) ÷ 4 which becomes eq.6a) 4×D = A + B + C + E
Hint #9
Substitute 3×F for A + B (from eq.2a), 0 for C, and 2×F for E in eq.6a: 4×D = 3×F + 0 + 2×F which becomes 4×D = 5×F Divide both sides of the above equation by 4: 4×D ÷ 4 = 5×F ÷ 4 which makes D = 1¼×F
Hint #10
Substitute 1¼×F for D in eq.4a: 2×F = 1¼×F + A Subtract 1¼×F from each side of the equation above: 2×F - 1¼×F = 1¼×F + A - 1¼×F which makes ¾×F = A
Hint #11
Substitute ¾×F for A in eq.2a: 3×F = ¾×F + B Subtract ¾×F from each side of the above equation: 3×F - ¾×F = ¾×F + B - ¾×F which makes 2¼×F = B
Solution
Substitute ¾×F for A, 2¼×F for B, 0 for C, 1¼×F for D, and 2×F for E in eq.1: ¾×F + 2¼×F + 0 + 1¼×F + 2×F + F = 29 which simplifies to 7¼×F = 29 Divide both sides of the above equation by 7¼: 7¼×F ÷ 7¼ = 29 ÷ 7¼ which means F = 4 making A = ¾×F = ¾ × 4 = 3 B = 2¼×F = 2¼ × 4 = 9 D = 1¼×F = 1¼ × 4 = 5 E = 2×F = 2 × 4 = 8 and ABCDEF = 390584