Puzzle for December 26, 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.
* AB, CD, and EF are 2-digit numbers (not A×B, C×D, or E×F).
Scratchpad
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Hint #1
In eq.3, replace A with D + F (from eq.2): D + F + B = D + E Subtract D from both sides of the equation above: D + F + B - D = D + E - D which becomes F + B = E which is equivalent to B + F = E In eq.5, replace B + F with E: eq.5a) A + C + D = E
Hint #2
In eq.4, replace E with A + C + D (from eq.5a): C + D = A + C + D - D which becomes C + D = A + C Subtract C from both sides of the above equation: C + D - C = A + C - C which makes D = A
Hint #3
In eq.2, substitute A for D: A = A + F Subtract A from each side: A - A = A + F - A which means 0 = F
Hint #4
Replace F with 0 in eq.5: A + C + D = B + 0 which is A + C + D = B In eq.5a, substitute B for A + C + D: B = E
Hint #5
Substitute A for D, and B for E in eq.4: C + A = B - A Subtract A from each side: C + A - A = B - A - A which becomes eq.4a) C = B - 2×A
Hint #6
eq.6 can be written as: A + 10×C + D = 10×E + F - (10×A + B) - D which is equivalent to A + 10×C + D = 10×E + F - 10×A - B - D Add 10×A + D to both sides of the above equation: A + 10×C + D + 10×A + D = 10×E + F - 10×A - B - D + 10×A + D which becomes eq.6a) 11×A + 10×C + 2×D = 10×E + F - B
Hint #7
Substitute (B - 2×A) for C (from eq.4a), A for D, B for E, and 0 for F in eq.6a: 11×A + 10×(B - 2×A) + 2×A = 10×B + 0 - B which is the same as 11×A + 10×B - 20×A + 2×A = 9×B which becomes 10×B - 7×A = 9×B Add (7×A - 9×B) to both sides: 10×B - 7×A + (7×A - 9×B) = 9×B + (7×A - 9×B) which simplifies to B = 7×A
Hint #8
Substitute 7×A for B in eq.4a: C = 7×A - 2×A = 5×A
Solution
Substitute 7×A for B and E, 5×A for C, A for D, and 0 for F in eq.1: A + 7×A + 5×A + A + 7×A + 0 = 21 which becomes 21×A = 21 Divide each side by 21: 21×A ÷ 21 = 21 ÷ 21 which means A = 1 making B = E = 7×A = 7 × 1 = 7 C = 5×A = 5 × 1 = 5 D = A = 1 and ABCDEF = 175170