Puzzle for February 26, 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 positive integer.
* "D mod B" equals the remainder of (D ÷ B).
Scratchpad
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Hint #1
In eq.4, replace D with A + E (from eq.1): B × E = A + E + E – A which becomes B × E = 2×E Divide both sides of the above equation by E: B × E ÷ E = 2×E ÷ E which makes B = 2
Hint #2
In eq.5, replace B with 2: eq.5a) D mod 2 = A Since the remainder of any positive integer divided by 2 must be < 2, then in eq.5a: D mod 2 < 2 which means D mod 2 = 0 or 1 In the above equation, replacing D mod 2 with A (from eq.5a) makes: A = 0 or 1 Since A is a positive integer, this means: A ≠ 0 which makes A = 1
Hint #3
In eq.1, substitute 1 for A: eq.1a) 1 + E = D
Hint #4
Substitute 1 for A, and 1 + E for D (from eq.1a) in eq.2: C + E = 1 + 1 + E which becomes C + E = 2 + E Subtract E from each side of the equation above: C + E – E = 2 + E – E which makes 2 = C
Hint #5
Substitute 2 for B and C, and 1 for A in eq.3: D – 2 = 1 + 2 + 2 + F which becomes D – 2 = 5 + F Add 2 to both sides of the above equation: D – 2 + 2 = 5 + F + 2 which makes eq.3a) D = 7 + F
Hint #6
In eq.5a, substitute (7 + F) for D (from eq.3a), and 1 for A: (7 + F) mod 2 = 1 If: 7 + F = even integer then: (7 + F) mod 2 = 0 ≠ 1 which means 7 + F ≠ even integer Therefore: 7 + F = odd integer which makes F = even integer
Hint #7
To make eq.3a true, check several possible values for F and D: If F = 2, then D = 7 + 2 = 9 If F = 4, then D = 7 + 4 = 11 If F > 4, then D > 11 Since D must be a one-digit positive integer, then: D = 9 which makes F = 2
Solution
Substitute 9 for D in eq.1a: 1 + E = 9 Subtract 1 from each side of the equation above: 1 + E – 1 = 9 – 1 which makes E = 8 and makes ABCDEF = 122982