Puzzle for March 2, 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, CD, and EF are 2-digit numbers (not B×C, C×D, or E×F).
Scratchpad
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Hint #1
Add B to each side of eq.5: E - B + B = A + B + B which becomes eq.5a) E = A + 2×B
Hint #2
In eq.2, replace E with A + 2×B (from eq.5a): A + B + C = A + 2×B + F Subtract both A and 2×B from each side of the above equation: A + B + C - A - 2×B = A + 2×B + F - A - 2×B which simplifies to eq.2a) C - B = F
Hint #3
In eq.4, replace F with C - B (from eq.2a): A + C + C - B = D - B Add B to each side of the above equation: A + C + C - B + B = D - B + B which becomes eq.4a) A + 2×C = D
Hint #4
In eq.3, substitute A + 2×C for D (from eq.4a): A + 2×C - C = B + E which becomes eq.3a) A + C = B + E eq.2 may be written as: B + A + C = E + F Substitute B + E for A + C (from eq.3a) into the equation above: B + B + E = E + F Subtract E from both sides: B + B + E - E = E + F - E which means 2×B = F
Hint #5
In eq.2a, replace F with 2×B: C - B = 2×B Add B to both sides: C - B + B = 2×B + B which makes C = 3×B
Hint #6
In eq.3, substitute 3×B for C: D - 3×B = B + E Add 3×B to each side: D - 3×B + 3×B = B + E + 3×B which becomes eq.3b) D = 4×B + E
Hint #7
eq.6 may be written as: 10×B + C + 10×C + D = 10×E + F which becomes 10×B + 11×C + D = 10×E + F Substitute 3×B for C, 4×B + E for D (from eq.3b), and 2×B for F in the above equation: 10×B + 11×3×B + 4×B + E = 10×E + 2×B which is equivalent to 10×B + 33×B + 4×B + E = 10×E + 2×B which becomes 47×B + E = 10×E + 2×B Subtract both 2×B and E from both sides: 47×B + E - 2×B - E = 10×E + 2×B - 2×B - E which becomes 45×B = 9×E Divide both sides by 9: 45×B ÷ 9 = 9×E ÷ 9 which means 5×B = E
Hint #8
Substitute 5×B for E in eq.3b: D = 4×B + 5×B which makes D = 9×B
Hint #9
Substitute 5×B for E in eq.5a: 5×B = A + 2×B Subtract 2×B from both sides: 5×B - 2×B = A + 2×B - 2×B which means 3×B = A
Solution
Substitute 3×B for A and C, 9×B for D, 5×B for E, and 2×B for F in eq.1: 3×B + B + 3×B + 9×B + 5×B + 2×B = 23 which simplifies to 23×B = 23 Divide both sides by 23: 23×B ÷ 23 = 23 ÷ 23 which makes B = 1 making A = C = 3×B = 3 × 1 = 3 D = 9×B = 9 × 1 = 9 E = 5×B = 5 × 1 = 5 F = 2×B = 2 × 1 = 2 and ABCDEF = 313952