Puzzle for June 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.
* AB, BC, DE, and EF are 2-digit numbers (not A×B, B×C, D×E, or E×F).
Scratchpad
Help Area
Hint #1
Add both C and D to each side of eq.3: B - C + C + D = A - D + C + D which becomes eq.3a) B + D = A + C Add both D and E to each side of eq.4: F - D + D + E = B - E + D + E which becomes eq.4a) F + E = B + D
Hint #2
eq.1 may be written as: A + C + B + D + F + E = 42 Replace A + C with B + D (from eq.3a), and F + E with B + D (from eq.4a): B + D + B + D + B + D = 42 which becomes 3×(B + D) = 42 Divide both sides of the above equation by 3: 3×(B + D) ÷ 3 = 42 ÷ 3 which makes eq.1a) B + D = 14
Hint #3
In eq.3a, replace B + D with 14 (from eq.1a): 14 = A + C Subtract A from each side of the above equation: 14 - A = A + C - A which becomes eq.3b) 14 - A = C In eq.4a, replace B + D with 14 (from eq.1a): F + E = 14 Subtract E from each side of the above equation: F + E - E = 14 - E which makes eq.4b) F = 14 - E
Hint #4
In eq.2, substitute 14 - A for C (from eq.3b), and 14 - E for F (from eq.4b): A + E = 14 - A + 14 - E Add both A and E to each side of the above equation: A + E + A + E = 14 - A + 14 - E + A + E which simplifies to 2×A + 2×E = 28 Divide both sides by 2: (2×A + 2×E) ÷ 2 = 28 ÷ 2 which becomes eq.2a) A + E = 14 Substitute A + E for 14 in eq.4b: F = A + E - E which makes F = A
Hint #5
Substitute A for F in eq.4b: A = 14 - E Substitute (14 - E) for A in eq.3b: 14 - (14 - E) = C which is the same as 14 - 14 + E = C which makes E = C
Hint #6
eq.5 may be written as: 10×A + B = C - D + 10×E + F which is equivalent to 10×A + B = C - D + 9×E + E + F which may also be written as 10×A + B = C - D + 9×E + F + E Substitute B + D for F + E (from eq.4a) in the above equation: 10×A + B = C - D + 9×E + B + D which becomes 10×A + B = C + 9×E + B Substitute E for C, and subtract B from both sides: 10×A + B - B = E + 9×E + B - B which becomes 10×A = 10×E Divide both sides by 10: 10×A ÷ 10 = 10×E ÷ 10 which makes A = E and which also makes F = A = E = C
Hint #7
Substitute A for E in eq.2a: A + A = 14 which makes 2×A = 14 Divide both sides by 2: 2×A ÷ 2 = 14 ÷ 2 which makes A = 7 and which also makes F = A = E = C = 7
Hint #8
eq.6 may be written as: B + 10×B + C = A + C + 10×D + E + E + F which becomes 11×B + C = A + C + 10×D + 2×E + F Subtract C from each side of the above equation: 11×B + C - C = A + C + 10×D + 2×E + F - C which becomes 11×B = A + 10×D + 2×E + F Substitute 7 for A, E, and F: 11×B = 7 + 10×D + 2×7 + 7 which becomes eq.6a) 11×B = 10×D + 28
Solution
Subtract B from both sides of eq.1a: B + D - B = 14 - B which makes eq.1b) D = 14 - B Substitute (14 - B) for D in eq.6a: 11×B = 10×(14 - B) + 28 which becomes 11×B = 140 - 10×B + 28 which becomes 11×B = 168 - 10×B Add 10×B to both sides: 11×B + 10×B = 168 - 10×B + 10×B which makes 21×B = 168 Divide both sides by 21: 21×B ÷ 21 = 168 ÷ 21 which means B = 8 making D = 14 - B = 14 - 8 = 6 (from eq.1b) and ABCDEF = 787677