Puzzle for December 9, 2018 ( )
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 DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
Add (D + C + F - B) to each side of eq.3: A + B - D + (D + C + F - B) = D - C - F + (D + C + F - B) which becomes eq.3a) A + C + F = 2×D - B Subtract the left and right sides of eq.4 from the respective sides of eq.3a: A + C + F - (A - B + C + F) = 2×D - B - (B + D) which is equivalent to A + C + F - A + B - C - F = 2×D - B - B - D which simplifies to B = D - 2×B Add 2×B to each side: B + 2×B = D - 2×B + 2×B which makes eq.3b) 3×B = D
Hint #2
Substitute 3×B for D in eq.4, and add B to each side: A - B + C + F + B = B + 3×B + B which becomes A + C + F = 5×B Substitute 5×B for A + C + F in eq.5: 5×B = B × F To make the above equation true, then: B = 0 and / or: F = 5
Hint #3
Check: B = 0 ... Substitute 0 for B in eq.3b: 3×0 = D which would make 0 = D Substitute 0 for B in eq.5: A + C + F = 0 × F which would make A + C + F = 0 Since A, C, and F must all be one-digit non-negative integers, then A = C = F = 0 Substitute 0 for A, B, C, D, and F in eq.1: 0 + 0 + 0 + 0 + E + 0 = 27 which would make E = 27 However, E must each be a one-digit non-negative integer, which means E ≠ 27 and means B ≠ 0 and therefore makes F = 5
Hint #4
eq.6 may be written as: A + 10×B + C + F = 10×D + E - (A + (10×B + C) + F) which is the same as A + 10×B + C + F = 10×D + E - A - 10×B - C - F Add A + C + F to both sides of the above equation: A + 10×B + C + F + A + C + F = 10×D + E - A - 10×B - C - F + A + C + F which becomes 2×A + 10×B + 2×C + 2×F = 10×D + E - 10×B Subtract 10×B from each side: 2×A + 10×B + 2×C + 2×F - 10×B = 10×D + E - 10×B - 10×B which becomes eq.6a) 2×A + 2×C + 2×F = 10×D + E - 20×B
Hint #5
From eq.3b, substitute (3×B) for D in eq.6a: 2×A + 2×C + 2×F = 10×(3×B) + E - 20×B which is equivalent to 2×A + 2×C + 2×F = 30×B + E - 20×B which becomes 2×A + 2×C + 2×F = 10×B + E which may be written as eq.6b) 2×(A + C + F) = 10×B + E
Hint #6
Substitute B × F for A + C + F (from eq.5) in eq.6b: 2×(B × F) = 10×B + E Subtract 10×B from each side: 2×(B × F) - 10×B = 10×B + E - 10×B which becomes 2×(B × F) - 10×B = E which may be written as (2×F - 10)×B = E Substitute 5 for F: (2×5 - 10)×B = E which becomes (10 - 10)×B = E which simplifies to 0×B = E which means 0 = E
Hint #7
In eq.2, replace E with 0: 0 = C - A Add A to each side: 0 + A = C - A + A which makes A = C
Hint #8
Substitute A for C, 3×B for D, 0 for E, and 5 for F in eq.1: A + B + A + 3×B + 0 + 5 = 27 which becomes 2×A + 4×B + 5 = 27 Subtract 5 from each side: 2×A + 4×B + 5 - 5 = 27 - 5 which becomes 2×A + 4×B = 22 Divide both sides by 2: (2×A + 4×B) ÷ 2 = 22 ÷ 2 which becomes A + 2×B = 11 Subtract 2×B from each side: A + 2×B - 2×B = 11 - 2×B which makes eq.1a) A = 11 - 2×B
Solution
Substitute (11 - 2×B) for both A (from eq.1a) and C, 3×B for D, and 5 for F in eq.3: (11 - 2×B) + B - 3×B = 3×B - (11 - 2×B) - 5 which becomes 11 - 4×B = 3×B - 11 + 2×B - 5 which becomes 11 - 4×B = 5×B - 16 Add 4×B + 16 to each side: 11 - 4×B + 4×B + 16 = 5×B - 16 + 4×B + 16 which simplifies to 27 = 9×B Divide both sides by 9: 27 ÷ 9 = 9×B ÷ 9 which makes 3 = B making C = A = 11 - 2×B = 11 - 2×3 = 11 - 6 = 5 (from eq.1a) D = 3×B = 3×3 = 9 (from eq.3b) and ABCDEF = 535905