Puzzle for April 29, 2022 ( )
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 D" equals the remainder of A ÷ D.
Scratchpad
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Hint #1
Subtract the left and right sides of eq.3 from the left and right sides of eq.2, respectively: D + F – (F – D) = B + E – (B + D – E) which becomes D + F – F + D = B + E – B – D + E which becomes 2×D = 2×E – D Add D to both sides of the above equation: 2×D + D = 2×E – D + D which makes 3×D = 2×E Divide both sides by 2: 3×D ÷ 2 = 2×E ÷ 2 which makes eq.2a) 1½×D = E
Hint #2
Since E and F are integers, then: E + F = an integer In eq.4, the above equation means: A ÷ D = an integer which means remainder A ÷ D = 0 which means A mod D = 0 In eq.5, replace A mod D with 0: 0 = B
Hint #3
In eq.2, replace B with 0, and E with 1½×D (from eq.2a): D + F = 0 + 1½×D which becomes D + F = 1½×D Subtract D from each side of the equation above: D + F – D = 1½×D – D which makes F = ½×D Multiply both sides by 2: 2 × F = 2 × ½×D which makes 2×F = D
Hint #4
In eq.2a, substitute (2×F) for D: 1½×(2×F) = E which makes 3×F = E
Hint #5
Substitute 2×F for D, and 3×F for E in eq.4: A ÷ 2×F = 3×F + F which becomes A ÷ 2×F = 4×F Multiply both sides of the above equation by (2×F): (2×F) × A ÷ 2×F = (2×F) × 4×F which makes eq.4a) A = 8×F²
Hint #6
Since F ≠ 0 (because D = 2×F, and F ≠ 0 (from eq.4 or eq.5)), check several positive values for A and F in eq.4a: If F = 1, then A = 8×F² = 8×1² = 8×1 = 8 If F = 2, then A = 8×F² = 8×2² = 8×4 = 32 If F > 2, then A > 32 Since A must be a one-digit integer, the above equations make: F = 1 and A = 8 making D = 2×F = 2 × 1 = 2 E = 3×F = 3 × 1 = 3
Solution
Substitute 8 for A, 0 for B, 2 for D, 1 for E, and 2 for F in eq.1: 8 + 0 + C + 2 + 3 + 1 = 22 which becomes 14 + C = 22 Subtract 14 from each side of the above equation: 14 + C – 14 = 22 – 14 which makes C = 8 and makes ABCDEF = 808231