Puzzle for April 7, 2019 ( )
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.
* BC and CD are 2-digit numbers (not B×C or C×D).
Scratchpad
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Hint #1
Add B + C to both sides of eq.4: B - C + B + C = F - B + E + B + C which becomes eq.4a) 2×B = F + E + C
Hint #2
Add E to both sides of eq.5: F - A - B + C + E = B - D + E which may be written as F + C + E - A - B = B - D + E In the above equation, replace F + C + E with 2×B (from eq.4a): 2×B - A - B = B - D + E which becomes B - A = B - D + E Subtract B from each side: B - A - B = B - D + E - B which becomes eq.5a) -A = -D + E
Hint #3
Add A + D to both sides of eq.5a: -A + A + D = -D + E + A + D which becomes eq.5b) D = E + A In eq.2, replace D with E + A: A + E + A = E + F Subtract E from each side: A + E + A - E = E + F - E which makes 2×A = F
Hint #4
Add C + D to both sides of eq.3: C - D + C + D = F - C + C + D which becomes eq.3a) 2×C = F + D Substitute 2×A for F, and E + A for D (from eq.5b) in the equation above: 2×C = 2×A + E + A which becomes eq.3b) 2×C = 3×A + E
Hint #5
eq.6 may be written as: 10×B + C - (10×C + D) = A - E + F which becomes 10×B + C - 10×C - D = A - E + F which may be written as 5×(2×B) - 9×C - D = A - E + F Substitute F + E + C for 2×B (from eq.4a) in the above equation: 5×(F + E + C) - 9×C - D = A - E + F which becomes 5×F + 5×E + 5×C - 9×C - D = A - E + F which becomes 5×F + 5×E - 4×C - D = A - E + F Add D + E to both sides of the above equation: 5×F + 5×E - 4×C - D + D + E = A - E + F + D + E which becomes eq.6a) 5×F + 6×E - 4×C = A + F + D
Hint #6
Substitute 2×C for F + D (from eq.3a), and (2×A) for F in eq.6a: 5×(2×A) + 6×E - 4×C = A + 2×C Add (4×C - A) to both sides: 5×(2×A) + 6×E - 4×C + (4×C - A) = A + 2×C + (4×C - A) which becomes 9×A + 6×E = 6×C which is equivalent to 9×A + 6×E = 3×(2×C)
Hint #7
Substitute 3×A + E for 2×C (from eq.3b) in the above equation: 9×A + 6×E = 3×(3×A + E) which becomes 9×A + 6×E = 9×A + 3×E Subtract (9×A + 3×E) from both sides: 9×A + 6×E - (9×A + 3×E) = 9×A + 3×E - (9×A + 3×E) which becomes 9×A + 6×E - 9×A - 3×E = 9×A + 3×E - 9×A - 3×E which simplifies to 3×E = 0 which means E = 0
Hint #8
Substitute 0 for E in eq.5b: D = 0 + A which means D = A
Hint #9
Substitute 0 for E in eq.3b: 2×C = 3×A + 0 which means 2×C = 3×A Divide both sides by 2: 2×C ÷ 2 = 3×A ÷ 2 which makes C = 1½×A
Hint #10
Substitute 2×A for F, 0 for E, and 1½×A for C in eq.4a: 2×B = 2×A + 0 + 1½×A which becomes 2×B = 3½×A Divide both sides by 2: 2×B ÷ 2 = 3½×A ÷ 2 which makes B = 1¾×A
Solution
Substitute 1¾×A for B, 1½×A for C, A for D, 0 for E, and 2×A for F in eq.1: A + 1¾×A + 1½×A + A + 0 + 2×A = 29 which simplifies to 7¼×A = 29 Divide both sides by 7¼: 7¼×A ÷ 7¼ = 29 ÷ 7¼ which means A = 4 making B = 1¾×A = 1¾ × 4 = 7 C = 1½×A = 1½ × 4 = 6 D = A = 4 F = 2×A = 2 × 1 = 8 and ABCDEF = 476408