Puzzle for July 2, 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.
* BC and DE are 2-digit numbers (not B×C or D×E).
** "F mod A" equals the remainder of (F ÷ A).
Scratchpad
Help Area
Hint #1
Add A to both sides of eq.3: D + F – A + A = A + B + A which becomes eq.3a) D + F = 2×A + B Add B and D to both sides of eq.1: F – B + B + D = B – D + B + D which becomes F + D = 2×B which may be written as eq.1a) D + F = 2×B
Hint #2
In eq.1a, replace D + F with 2×A + B (from eq.3a): 2×A + B = 2×B Subtract B from both sides of the above equation: 2×A + B – B = 2×B – B which makes 2×A = B
Hint #3
In eq.4, replace B with 2×A: 2×A ÷ A = F – D which becomes 2 = F – D Add D to both sides of the equation above: 2 + D = F – D + D which becomes eq.4a) 2 + D = F
Hint #4
In eq.3, substitute 2 + D for F (from eq.4a), and 2×A for B: D + 2 + D – A = A + 2×A which becomes 2×D + 2 – A = 3×A In the above equation, subtract 2 from both sides, and add A to both sides: 2×D + 2 – A – 2 + A = 3×A – 2 + A which becomes 2×D = 4×A – 2 Divide both sides by 2: 2×D ÷ 2 = (4×A – 2) ÷ 2 which becomes eq.3a) D = 2×A – 1
Hint #5
Substitute 2×A – 1 for D (from eq.3a) into eq.4a: 2 + 2×A – 1 = F which becomes eq.4b) 2×A + 1 = F
Hint #6
Substitute (2×A – 1) for D (from eq.3a), and (2×A + 1) for F (from eq.4b) in eq.2: A – (2×A – 1) = (2×A – 1) – (2×A + 1) which becomes A – 2×A + 1 = 2×A – 1 – 2×A – 1 which becomes –A + 1 = –2 Add A and 2 to both sides of the above equation: –A + 1 + A + 2 = –2 + A + 2 which makes 3 = A making B = 2×A = 2×3 = 6 D = 2×A – 1 = 2×3 – 1 = 6 – 1 = 5 (from eq.3a) F = 2×A + 1 = 2×3 + 1 = 6 + 1 = 7 (from eq.4b)
Hint #7
Substitute 7 for F, 3 for A, 6 for B, and 5 for D in eq.6: 7 mod 3 = 6 – 5 + E which becomes remainder of (7 ÷ 3) = 1 + E which becomes 1 = 1 + E Subtract 1 from each side of the above equation: 1 – 1 = 1 + E – 1 which makes 0 = E
Solution
eq.5 may be written as: 10×B + C – A = 10×D + E + F Substitute 6 for B, 3 for A, 5 for D, 0 for E, and 7 for F in the equation above: 10×6 + C – 3 = 10×5 + 0 + 7 which becomes 60 + C – 3 = 50 + 7 which becomes 57 + C = 57 Subtract 57 from each side of the equation above: 57 + C – 57 = 57 – 57 which makes C = 0 and makes ABCDEF = 360507