Puzzle for February 9, 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.
* CD and DE are 2-digit numbers (not C×D or D×E).
Scratchpad
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Hint #1
Add B + F to each side of eq.4: C - B + B + F = E - F + B + F which becomes C + F = E + B which may be written as C + F = B + E In eq.3, replace B + E with C + F: C + F = C + D Subtract C from each side of the above equation: C + F - C = C + D - C which makes F = D
Hint #2
eq.6 may be written as: 10×D + E - (10×C + D) = E - C which is the same as 10×D + E - 10×C - D = E - C which becomes 9×D + E - 10×C = E - C Subtract E from both sides, and add 10×C to both sides of the above equation: 9×D + E - 10×C - E + 10×C = E - C - E + 10×C which simplifies to 9×D = 9×C Divide both sides by 9: 9×D ÷ 9 = 9×C ÷ 9 which makes D = C
Hint #3
eq.2 may be written as: A + B = E + F Add the left and right sides of the above equation to the left and right sides of eq.4, respectively: C - B + A + B = E - F + E + F which becomes eq.4a) C + A = 2×E
Hint #4
In eq.3, substitute C for D: B + E = C + C which becomes B + E = 2×C Multiply both sides of the above equation by 2: 2×(B + E) = 4×C which is the same as eq.3a) 2×B + 2×E = 4×C
Hint #5
In eq.5, replace D with C: C + C - A = A - B - C Add both A and C to each side of the above equation: C + C - A + A + C = A - B - C + A + C which simplifies to eq.5a) 3×C = 2×A - B
Hint #6
Substitute C + A for 2×E (from eq.4a) in eq.3a: 2×B + C + A = 4×C Subtract C from each side of the above equation: 2×B + C + A - C = 4×C - C which becomes 2×B + A = 3×C Substitute 2×A - B for 3×C (from eq.5a): 2×B + A = 2×A - B Add (B - A) to both sides: 2×B + A + (B - A) = 2×A - B + (B - A) which simplifies to 3×B = A
Hint #7
Substitute (3×B) for A in eq.5a: 3×C = 2×(3×B) - B which becomes 3×C = 6×B - B which makes 3×C = 5×B Divide both sides of the above equation by 3: 3×C ÷ 3 = 5×B ÷ 3 which makes C = 1⅔×B
Hint #8
Substitute 3×B for A, and 1⅔×B for C in eq.4a: 1⅔×B + 3×B = 2×E which means 4⅔×B = 2×E Divide both sides by 2: 4⅔×B ÷ 2 = 2×E ÷ 2 which makes 2⅓×B = E
Solution
Substitute 3×B for A, 1⅔×B for C and D and F, and 2⅓×B for E in eq.1: 3×B + B + 1⅔×B + 1⅔×B + 2⅓×B + 1⅔×B = 34 which becomes 11⅓×B = 34 Divide both sides by 11⅓: 11⅓×B ÷ 11⅓ = 34 ÷ 11⅓ = 3 which makes B = 3 making A = 3×B = 3 × 3 = 9 C = D = F = 1⅔×B = 1⅔ × 3 = 5 E = 2⅓×B = 2⅓ × 3 = 7 and ABCDEF = 935575