Puzzle for December 6, 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.
Scratchpad
Help Area
Hint #1
In eq.5, replace C with B + F (from eq.6): B + F – E = B – F Subtract B from each side of the above equation: B + F – E – B = B – F – B which becomes F – E = –F Add both E and F to each side: F – E + E + F = –F + E + F which makes 2×F = E
Hint #2
In eq.3, replace E with 2×F: C – A = A – D + 2×F Add A and D to each side of the above equation: C – A + A + D = A – D + 2×F + A + D which becomes C + D = 2×A + 2×F which may be written as eq.3a) C + D = 2×(A + F)
Hint #3
In eq.3a, substitute B – A for A + F (from eq.4): C + D = 2×(B – A) which may be written as eq.3b) C + D = 2×B – 2×A
Hint #4
In eq.4, add A to both sides, and subtract F from each side: B – A + A – F = A + F + A – F which becomes B – F = 2×A Substitute (B – F) for 2×A in eq.3b: C + D = 2×B – (B – F) which becomes eq.3c) C + D = B + F
Hint #5
Substitute C for B + F (from eq.6) in eq.3c: C + D = C Subtract C from each side of the equation above: C + D – C = C – C which makes D = 0
Hint #6
Add A to both sides of eq.4: B – A + A = A + F + A which becomes B = 2×A + F Substitute 0 for D, and 2×F for E, and 2×A + F for B in eq.2: 0 + 2×F = A + 2×A + F which becomes 2×F = 3×A + F Subtract F from each side of the equation above: 2×F – F = 3×A + F – F which makes F = 3×A and which also makes E = 2×F = 2×(3×A) = 6×A
Hint #7
Substitute 0 for D, and 6×A for E in eq.2: 0 + 6×A = A + B Subtract A from each side of the above equation: 0 + 6×A – A = A + B – A which makes 5×A = B
Hint #8
Substitute 5×A for B, and 3×A for F in eq.6: 5×A + 3×A = C which makes 8×A = C
Solution
Substitute 5×A for B, 8×A for C, 0 for D, 6×A for E, and 3×A for F in eq.1: A + 5×A + 8×A + 0 + 6×A + 3×A = 23 which simplifies to 23×A = 23 Divide both sides of the equation above by 23: 23×A ÷ 23 = 23 ÷ 23 which means A = 1 making B = 5×A = 5 × 1 = 5 C = 8×A = 8 × 1 = 8 E = 6×A = 6 × 1 = 6 F = 3×A = 3 × 1 = 3 and ABCDEF = 158063