Puzzle for February 23, 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.
* EF is a 2-digit number (not E×F).
Our thanks go out to Judah S (age 14) for submitting this puzzle. Thank you, Judah!
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Hint #1
eq.5 may be written as: 10×E + F = A + C In the above equation, replace F with A – E (from eq.3): 10×E + A – E = A + C which becomes 9×E + A = A + C Subtract A from both sides of the above equation: 9×E + A – A = A + C – A which makes 9×E = C
Hint #2
In eq.2, replace C with 9×E: 9×E = D + E Subtract E from both sides of the equation above: 9×E – E = D + E – E which makes 8×E = D
Hint #3
In eq.4, substitute 8×E for D, and 9×E for C: B + 8×E = 9×E + E which becomes B + 8×E = 10×E Subtract 8×E from each side of the equation above: B + 8×E – 8×E = 10×E – 8×E which makes B = 2×E
Hint #4
Substitute 8×E for D, 2×E for B, and 9×E for C in eq.6: 8×E ÷ 2×E = 9×E + E – F which becomes 4 = 10×E – F In the above equation, add F to both sides, and subtract 4 from both sides: 4 + F – 4 = 10×E – F + F – 4 which becomes eq.6a) F = 10×E – 4
Hint #5
Substitute 10×E – 4 for F (from eq.6a) in eq.3: 10×E – 4 = A – E Add E to both sides of the equation above: 10×E – 4 + E = A – E + E which becomes eq.3a) 11×E – 4 = A
Solution
In eq.1, substitute 11×E – 4 for A (from eq.3a), 2×E for B, 9×E for C, 8×E for D, and 10×E – 4 for F (from eq.6a): 11×E – 4 + 2×E + 9×E + 8×E + E + 10×E – 4 = 33 which simplifies to 41×E – 8 = 33 Add 8 to both sides of the above equation: 41×E – 8 + 8 = 33 + 8 which makes 41×E = 41 Divide both sides by 41: 41×E ÷ 41 = 41 ÷ 41 which means E = 1 making A = 11×E – 4 = 11×1 – 4 = 11 – 4 = 7 (from eq.3a) B = 2×E = 2 × 1 = 2 C = 9×E = 9 × 1 = 9 D = 8×E = 8 × 1 = 8 F = 10×E – 4 = 10×1 – 4 = 10 – 4 = 6 (from eq.6a) and ABCDEF = 729816