Puzzle for December 7, 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, BC, and CD are 2-digit numbers (not A×B, B×C, or C×D).
Scratchpad
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Hint #1
Add B to each side of eq.4: C - B + B = A + B + B which becomes eq.4a) C = A + 2×B
Hint #2
eq.6 may be re-written as: 10×A + B + 10×B + C + D = 10×C + D which becomes 10×A + 11×B + C + D = 10×C + D Subtract both C and D from each side of the equation above: 10×A + 11×B + C + D - C - D = 10×C + D - C - D which becomes eq.6a) 10×A + 11×B = 9×C
Hint #3
From eq.4a, substitute (A + 2×B) for C in eq.6a: 10×A + 11×B = 9×(A + 2×B) which is the same as 10×A + 11×B = 9×A + 18×B Subtract (9×A + 11×B) from each side: 10×A + 11×B - (9×A + 11×B) = 9×A + 18×B - (9×A + 11×B) which may be written as 10×A + 11×B - 9×A - 11×B = 9×A + 18×B - 9×A - 11×B which simplifies to A = 7×B
Hint #4
Substitute 7×B for A in eq.4a: C = 7×B + 2×B which means C = 9×B
Hint #5
Substitute 7×B for A in eq.3: 7×B - B = D - E which becomes 6×B = D - E Add E to each side: 6×B + E = D - E + E which becomes eq.3a) 6×B + E = D
Hint #6
Substitute 7×B for A, 9×B for C, and 6×B + E for D (from eq.3a) in eq.5: 7×B + B + F = 9×B + 6×B + E which becomes 8×B + F = 15×B + E Subtract 8×B from each side: 8×B + F - 8×B = 15×B + E - 8×B which becomes eq.5a) F = 7×B + E
Hint #7
Substitute 7×B for A, and 7×B + E for F (from eq.5a) in eq.2: 7×B = E + 7×B + E Subtract 7×B from both sides: 7×B - 7×B = E + 7×B + E - 7×B which becomes 0 = 2×E which means 0 = E
Hint #8
In eq.3a, replace E with 0: 6×B + 0 = D which makes 6×B = D
Hint #9
Replace E with 0 in eq.5a: F = 7×B + 0 = 7×B
Solution
Substitute 7×B for A and F, 9×B for C, 6×B for D, and 0 for E in eq.1: 7×B + B + 9×B + 6×B + 0 + 7×B = 30 which simplifies to 30×B = 30 Divide both sides by 30: 30×B ÷ 30 = 30 ÷ 30 which means B = 1 making A = F = 7×B = 7 × 1 = 7 C = 9×B = 9 × 1 = 9 D = 6×B = 6 × 1 = 6 and ABCDEF = 719607