Puzzle for March 13, 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.
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Hint #1
Since D ≠ 0 (from eq.4), divide both sides of eq.1 by D: F ÷ D = (A + B) × D ÷ D which means F ÷ D = A + B In eq.4, replace (F ÷ D) with A + B, and D + E with A – E (from eq.2): B – C + A + B = C + A – E which becomes 2×B – C + A = C + A – E In the above equation, subtract A from both sides, and add C to both sides: 2×B – C + A – A + C = C + A – E – A + C which becomes eq.4a) 2×B = 2×C – E
Hint #2
Add D to both sides of eq.3: C + D + D = E – D + D which becomes eq.3a) C + 2×D = E In eq.4a, substitute (C + 2×D) for E (from eq.3a): 2×B = 2×C – (C + 2×D) which is equivalent to 2×B = 2×C – C – 2×D which becomes 2×B = C – 2×D Add 2×D to both sides of the above equation: 2×B + 2×D = C – 2×D + 2×D which becomes eq.4b) 2×B + 2×D = C
Hint #3
In eq.3a, replace C with 2×B + 2×D (from eq.4b): 2×B + 2×D + 2×D = E which becomes eq.3b) 2×B + 4×D = E
Hint #4
Add E to both sides of eq.2: D + E + E = A – E + E which becomes D + 2×E = A In the above equation, substitute (2×B + 4×D) for E (from eq.3b): D + 2×(2×B + 4×D) = A which becomes D + 4×B + 8×D = A which becomes eq.2a) 4×B + 9×D = A
Hint #5
To make eq.2a true, check several possible values for D, B, and A: If D = 1 then A = 4×B + 9×1 = 4×B + 9 making B ≤ 0 (since 0 ≤ A ≤ 9) If D ≥ 2 then A ≥ 4×B + 9×2 = 4×B + 18 making B ≤ –3 (since 0 ≤ A ≤ 9) Since B must be a non-negative integer, the above equations and inequalities make: B = 0 which makes D = 1 and also makes A = 4×B + 9×D = 4×0 + 9×1 = 0 + 9 = 9 (from eq.2a)
Hint #6
Substitute 9 for A, 0 for B, and 1 for D in eq.1: F = (9 + 0) × 1 which becomes F = 9 × 1 which makes F = 9
Hint #7
Substitute 0 for B, and 1 for D in eq.4b: 2×0 + 2×1 = C which makes 2 = C
Solution
Substitute 2 for C, and 1 for D in eq.3a: 2 + 2×1 = E which becomes 2 + 2 = E which makes 4 = E and makes ABCDEF = 902149