Puzzle for July 28, 2021 ( )
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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.
* AB is a 2-digit number (not A×B).
Our thanks go out to Judah S (age 14) for sending us this puzzle. Thanks, Judah!
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Hint #1
In eq.4, substitute (B × F) for C (from eq.3): B ÷ (B × F) = A which becomes 1 ÷ F = A Multiply both sides of the equation above by F: (1 ÷ F) × F = A × F eq.4a) 1 = A × F A and F are non-negative integers. The only two non-negative integers for A and F that can make eq.4a true are: A = 1 and F = 1
Hint #2
eq.5 may be written as: (10×A + B) ÷ E = A + F In the above equation, replace A with 1, and F with 1: (10×1 + B) ÷ E = 1 + 1 which becomes (10 + B) ÷ E = 2 Multiply both sides by E: ((10 + B) ÷ E) × E = 2 × E which becomes 10 + B = 2×E Divide both sides by 2: (10 + B) ÷ 2 = 2×E ÷ 2 which makes eq.5a) 5 + ½×B = E
Hint #3
In eq.2, replace E with 5 + ½×B (from eq.5a), and replace A and F with 1: B + 5 + ½×B = 1 + D + 1 which becomes 1½×B + 5 = D + 2 Subtract 2 from both sides of the equation above: 1½×B + 5 – 2 = D + 2 – 2 which makes eq.2a) 1½×B + 3 = D
Hint #4
In eq.3, substitute 1 for F: C = B × 1 which makes C = B
Solution
Substitute 1 for A and F, B for C, 1½×B + 3 for D (from eq.2a), and 5 + ½×B for E (from eq.5a) in eq.1: 1 + B + B + 1½×B + 3 + 5 + ½×B + 1 = 26 which simplifies to 10 + 4×B = 26 Subtract 10 from each side of the equation above: 10 + 4×B – 10 = 26 – 10 which makes 4×B = 16 Divide both sides by 4: 4×B ÷ 4 = 16 ÷ 4 which means B = 4 making C = B = 4 D = 1½×B + 3 = 1½×4 + 3 = 6 + 3 = 9 (from eq.2a) E = 5 + ½×B = 5 + ½×4 = 5 + 2 = 7 (from eq.5a) and ABCDEF = 144971