Puzzle for July 15, 2021 ( )
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.
* "A mod B" is the remainder of (A ÷ B).
Scratchpad
Help Area
Hint #1
eq.5 may be written as: D = (C + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (C + E + F) ÷ 3 which becomes eq.5a) 3×D = C + E + F
Hint #2
Add E to both sides of eq.2: D + E + E = C + F + E which becomes D + 2×E = C + E + F In the above equation, replace 3×D for C + E + F (from eq.5a): D + 2×E = 3×D Subtract D from both sides: D + 2×E – D = 3×D – D which becomes 2×E = 2×D Divide both sides by 2: 2×E ÷ 2 = 2×D ÷ 2 which makes E = D
Hint #3
Add B and F to both sides of eq.4: B + E – F + B + F = A – B + B + F which becomes 2×B + E = A + F In the above equation, substitute D for E, and C + D for A + F (from eq.3): 2×B + D = C + D Subtract D from each side: 2×B + D – D = C + D – D which makes 2×B = C
Hint #4
Substitute 2×B for C in eq.1: F = B + 2×B which makes F = 3×B
Hint #5
Substitute D for E, 2×B for C, and 3×B for F in eq.2: D + D = 2×B + 3×B which makes 2×D = 5×B Divide both sides of the equation above by 2: 2×D ÷ 2 = 5×B ÷ 2 which makes D = 2½×B and also makes E = D = 2½×B
Hint #6
Substitute 2×B for C, 2½×B for D, and 3×B for F in eq.3: 2×B + 2½×B = A + 3×B which becomes 4½×B = A + 3×B Subtract 3×B from each side of the above equation: 4½×B – 3×B = A + 3×B – 3×B which makes 1½×B = A
Solution
Substitute 1½×B for A, and D for E in eq.6: 1½×B mod B = D ÷ D which means remainder of (1½×B ÷ B) = 1 which means ½×B = 1 Multiply both sides of the above equation by 2: 2 × ½×B = 2 × 1 which makes B = 2 making A = 1½×B = 1½ × 2 = 3 C = 2×B = 2 × 2 = 4 D = E = 2½×B = 2½ × 2 = 5 F = 3×B = 3 × 2 = 6 and ABCDEF = 324556