Puzzle for January 27, 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.
* AB, BC, DE, and EF are 2-digit numbers (not A×B, B×C, D×E, or E×F).
Scratchpad
Help Area
Hint #1
eq.5 may be written as: 10×A + B = C + 10×D + E Subtract E from both sides of the above equation: 10×A + B - E = C + 10×D + E - E which becomes eq.5a) 10×A + B - E = C + 10×D
Hint #2
In eq.5a, replace B - E with C (from (eq.3): 10×A + C = C + 10×D Subtract C from each side of the above equation: 10×A + C - C = C + 10×D - C which becomes 10×A = 10×D Divide both sides by 10: 10×A ÷ 10 = 10×D ÷ 10 which makes A = D
Hint #3
In eq.3, substitute (B + D - E) for E (from eq.4): B - (B + D - E) = C which is the same as B - B - D + E = C which becomes -D + E = C which may also be written as E - D = C In eq.2, substitute (E - D) for the first C, and (B - E) for the second C (from eq.3): A + (E - D) + D - F = B - (B - E) which is the same as A + E - D + D - F = B - B + E which becomes A + E - F = E Add (F - E) to each side: A + E - F + (F - E) = E + (F - E) which simplifies to A = F
Hint #4
In eq.4, replace D with A: B + A - E = E Add E to each side of the above equation: B + A - E + E = E + E which is the same as A + B = 2×E Divide both sides by 2: (A + B) ÷ 2 = 2×E ÷ 2 which means eq.4a) ½×A + ½×B = E
Hint #5
eq.6 may be written as: 10×E + F = 10×B + C - E Add E to both sides of the equation above: 10×E + F + E = 10×B + C - E + E which becomes 11×E + F = 10×B + C Substitute (½×A + ½×B) for E (from eq.4a), and A for F: 11×(½×A + ½×B) + A = 10×B + C which becomes 5½×A + 5½×B + A = 10×B + C Subtract 10×B from each side: 5½×A + 5½×B + A - 10×B = 10×B + C - 10×B which simplifies to eq.6a) 6½×A - 4½×B = C
Hint #6
Substitute (½×A + ½×B) for E (from eq.4a), and 6½×A - 4½×B for C (from eq.6a) in eq.3: B - (½×A + ½×B) = 6½×A - 4½×B which is equivalent to B - ½×A - ½×B = 6½×A - 4½B which becomes ½×B - ½×A = 6½×A - 4½×B Add ½×A + 4½×B to both sides: ½×B - ½×A + ½×A + 4½×B = 6½×A - 4½×B + ½×A + 4½×B which simplifies to 5×B = 7×A Divide both sides by 5: 5×B ÷ 5 = 7×A ÷ 5 which makes B = 1⅖×A
Hint #7
In eq.6a, replace B with 1⅖×A: 6½×A - 4½×(1⅖×A) = C which may be written as 6.5×A - 4.5×(1.4×A) = C which becomes 6.5×A - 6.3×A = C which becomes 0.2×A = C which is the same as ⅕×A = C
Hint #8
In eq.4a, substitute 1⅖×A for B: ½×A + ½×(1⅖×A) = E which may be written as 0.5×A + 0.5×(1.4×A) = E which becomes 0.5×A + 0.7×A = E which becomes 1.2×A = E which is the same as 1⅕×A = E
Solution
Substitute A for D and F, 1⅖×A for B, ⅕×A for C, and 1⅕×A for E in eq.1: A + 1⅖×A + ⅕×A + A + 1⅕×A + A = 29 which becomes 5⅘×A = 29 Divide both sides by 5⅘: 5⅘×A ÷ 5⅘ = 29 ÷ 5⅘ which makes A = 5 making B = 1⅖×A = 1⅖ × 5 = 7 C = ⅕×A = ⅕ × 5 = 1 D = F = A = 5 E = 1⅕×A = 1⅕ × 5 = 6 and ABCDEF = 571565