Puzzle for February 18, 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.
Scratchpad
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Hint #1
Add E to both sides of eq.2: B – E + E = A + E which becomes B = A + E In eq.3, replace B with (A + E): C – (A + E) = A – E which becomes C – A – E = A – E Add A and E to both sides of the equation above: C – A – E + A + E = A – E + A + E which makes eq.2a) C = 2×A
Hint #2
In eq.5, subtract E from both sides, and add A and F to both sides: E – A – C – E + A + F = A – F – E + A + F which becomes –C + F = 2×A – E which may be written as F – C = 2×A – E In eq.4, replace F – C with 2×A – E: D + 2×A – E = A + C Subtract A from each side of the above equation: D + 2×A – E – A = A + C – A which becomes eq.5a) D + A – E = C
Hint #3
In eq.5a, substitute C – B for A – E (from eq.3): D + C – B = C In the above equation, subtract C from both sides, and add B to both sides: D + C – B – C + B = C – C + B which makes D = B
Hint #4
Substitute 2×A for C (from eq.2a) in eq.5: E – A – 2×A = A – F which becomes E – 3×A = A – F Add 3×A and F to both sides of the above equation: E – 3×A + 3×A + F = A – F + 3×A + F which becomes eq.5b) E + F = 4×A
Hint #5
eq.6 may be written as: F – B – D = A – (E + F) Substitute D for B, and 4×A for E + F (from eq.5b) in the equation above: F – D – D = A – (4×A) which becomes F – 2×D = –3×A Add 2×D to both sides: F – 2×D + 2×D = –3×A + 2×D which becomes F = –3×A + 2×D which may be written as eq.6a) F = 2×D – 3×A
Hint #6
Substitute 2×D – 3×A for F (from eq.6a), and 2×A for C (from eq.2a) in eq.4: D + 2×D – 3×A – 2×A = A + 2×A which becomes 3×D – 5×A = 3×A Add 5×A to both sides of the above equation: 3×D – 5×A + 5×A = 3×A + 5×A which becomes 3×D = 8×A Divide both sides by 8: 3×D ÷ 8 = 8×A ÷ 8 which makes ⅜×D = A
Hint #7
Substitute (⅜×D) for A in eq.2a: C = 2×(⅜×D) which makes C = ¾×D
Hint #8
Substitute (⅜×D) for A in eq.6a: F = 2×D – 3×(⅜×D) which becomes F = 2×D – 1⅛×D which makes F = ⅞×D
Hint #9
Substitute ⅞×D for F, and (⅜×D) for A in eq.5b: E + ⅞×D = 4×(⅜×D) which becomes E + ⅞×D = 1½×D Subtract ⅞×D from each side of the equation above: E + ⅞×D – ⅞×D = 1½×D – ⅞×D which makes E = ⅝×D
Solution
Substitute ⅜×D for A, D for B, ¾×D for C, ⅝×D for E, and ⅞×D for F in eq.1: ⅜×D + D + ¾×D + D + ⅝×D + ⅞×D = 37 which simplifies to 4⅝×D = 37 Divide both sides of the above equation by 4⅝: 4⅝×D ÷ 4⅝ = 37 ÷ 4⅝ which means D = 8 making A = ⅜×D = ⅜ × 8 = 3 B = D = 8 C = ¾×D = ¾ × 8 = 6 E = ⅝×D = ⅝ × 8 = 5 F = ⅞×D = ⅞ × 8 = 7 and ABCDEF = 386857