Puzzle for September 21, 2019 ( )
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 A and C to both sides of eq.4: E – C + A + C = F – A + A + C which becomes E + A = F + C which is the same as E + A = C + F In eq.2, replace C + F with E + A: B + E = E + A Subtract E from both sides of the above equation: B + E – E = E + A – E which makes B = A
Hint #2
In eq.3, replace B with A: A = A + D Subtract A from both sides of the equation above: A – A = A + D – A which makes 0 = D
Hint #3
In eq.5, substitute 0 for D: 0 – C = C – 0 – E – F Add C, E, and F to both sides of the above equation: 0 – C + C + E + F = C – 0 – E – F + C + E + F which simplifies to eq.5a) E + F = 2×C
Hint #4
Substitute 0 for D, and A for B in eq.6: C + 0 + E – A = A + A C + E – A = 2×A In the above equation, add A to both sides, and subtract C from both sides: C + E – A + A – C = 2×A + A – C which becomes eq.6a) E = 3×A – C
Hint #5
Substitute 3×A – C for E (from eq.6a) in eq.5a: 3×A – C + F = 2×C Add C to both sides of the above equation: 3×A – C + F + C = 2×C + C which becomes 3×A + F = 3×C Subtract 3×A from both sides of the equation above: 3×A + F – 3×A = 3×C – 3×A which becomes eq.5b) F = 3×C – 3×A
Hint #6
Substitute 3×A – C for E (from eq.6a), and 3×C – 3×A for F (from eq.5b) in eq.4: 3×A – C – C = 3×C – 3×A – A which becomes 3×A – 2×C = 3×C – 4×A Add 2×C and 4×A to both sides: 3×A – 2×C + 2×C + 4×A = 3×C – 4×A + 2×C + 4×A which simplifies to eq.4a) 7×A = 5×C
Hint #7
In eq.6a, add C to both sides, and subtract E from both sides: E + C – E = 3×A – C + C – E which becomes eq.6b) C = 3×A – E Substitute (3×A – E) for C (from eq.6b) in eq.4a: 7×A = 5×(3×A – E) which is equivalent to 7×A = 15×A – 5×E In the above equation, add 5×E to each side, and subtract 7×A from both sides: 7×A + 5×E – 7×A = 15×A – 5×E + 5×E – 7×A which simplifies to 5×E = 8×A Divide both sides by 8: 5×E ÷ 8 = 8×A ÷ 8 which makes ⅝×E = A and also makes B = A = ⅝×E
Hint #8
Substitute (⅝×E) for A in eq.6b: C = 3×(⅝×E) – E which becomes C = 1⅞×E – E which means C = ⅞×E
Hint #9
Substitute (⅞×E) for C, and (⅝×E) for A in eq.5b: F = 3×(⅞×E) – 3×(⅝×E) which becomes F = 2⅝×E – 1⅞×E which means F = ¾×E
Solution
Substitute ⅝×E for A and B, ⅞×E for C, 0 for D, and ¾×E for F in eq.1: ⅝×E + ⅝×E + ⅞×E + 0 + E + ¾×E = 31 which simplifies to 3⅞×E = 31 Divide both sides of the equation above by 3⅞: 3⅞×E ÷ 3⅞ = 31 ÷ 3⅞ which means E = 8 making A = B = ⅝×E = ⅝ × 8 = 5 C = ⅞×E = ⅞ × 8 = 7 F = ¾×E = ¾ × 8 = 6 and ABCDEF = 557086