Puzzle for February 16, 2020 ( )
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.
* AB and DE are 2-digit numbers (not A×B or D×E).
Scratchpad
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Hint #1
Subtract C from both sides of eq.2: C + D – C = B + E – C which becomes D = B + E – C In eq.3, replace D with B + E – C: B + E – C = E + F – B In the equation above, add B and C to both sides, and subtract E from both sides: B + E – C + B + C – E = E + F – B + B + C – E which simplifies to 2×B = F + C which may be written as eq.3a) 2×B = C + F
Hint #2
In eq.4, replace C + F with 2×B (from eq.3a): A + 2×B – E = D + E In the equation above, add E to each side, and subtract A from both sides: A + 2×B – E + E – A = D + E + E – A which becomes eq.4a) 2×B = D + 2×E – A
Hint #3
In eq.5, replace C + F with 2×B (from eq.3a): B + 2×B – A = D + E + A Add A to both sides of the equation above: B + 2×B – A + A = D + E + A + A which becomes eq.5a) 3×B = D + E + 2×A
Hint #4
Multiply both sides of eq.4a by 3: 3 × 2×B = 3 × (D + 2×E – A) which becomes eq.4b) 6×B = 3×D + 6×E – 3×A Multiply both sides of eq.5a by 2: 2 × 3×B = 2 × (D + E + 2×A) which becomes eq.5b) 6×B = 2×D + 2×E + 4×A In eq.4b, substitute 2×D + 2×E + 4×A for 6×B (from eq.5b): 2×D + 2×E + 4×A = 3×D + 6×E – 3×A In the equation above, add 3×A to each side, and subtract both 2×D and 2×E from each side: 2×D + 2×E + 4×A + 3×A – 2×D – 2×E = 3×D + 6×E – 3×A + 3×A – 2×D – 2×E which simplifies to eq.5c) 7×A = D + 4×E
Hint #5
eq.6 may be written as: 10×A + B + D + E = 10×D + E – A Subtract 10×A, D, and E from each side of the equation above: 10×A + B + D + E – 10×A – D – E = 10×D + E – A – 10×A – D – E which simplifies to B = 9×D – 11×A Substitute (9×D – 11×A) for B in eq.4a: 2×(9×D – 11×A) = D + 2×E – A which is equivalent to 18×D – 22×A = D + 2×E – A In the above equation, add A to each side, and subtract D from each side: 18×D – 22×A + A – D = D + 2×E – A + A – D which becomes eq.6a) 17×D – 21×A = 2×E
Hint #6
eq.5c may be written as: 7×A = D + 2×(2×E) Substitute 17×D – 21×A for 2×E (from eq.6a) in the above equation: 7×A = D + 2×(17×D – 21×A) which becomes 7×A = D + 34×D – 42×A Add 42×A to both sides: 7×A + 42×A = D + 34×D – 42×A + 42×A which means 49×A = 35×D Divide both sides by 7: 49×A ÷ 7 = 35×D ÷ 7 which makes 7×A = 5×D
Hint #7
eq.6a may be written as: 17×D – 3×(7×A) = 2×E Substitute 5×D for 7×A in the above equation: 17×D – 3×(5×D) = 2×E which becomes 17×D – 15×D = 2×E which means 2×D = 2×E Divide both sides by 2: 2×D ÷ 2 = 2×E ÷ 2 which makes D = E
Hint #8
Substitute D for E in eq.2: C + D = B + D Subtract D from both sides of the above equation: C + D – D = B + D – D which makes C = B
Hint #9
Substitute D for E in eq.3: D = D + F – B In the above equation, subtract D from both sides, and add B to each side: D – D + B = D + F – B – D + B which makes B = F
Hint #10
Substitute D for E in eq.4a: 2×B = D + 2×D – A which becomes eq.4c) 2×B = 3×D – A Multiply both sides of eq.4c by 7: 7 × 2×B = 7 × (3×D – A) which becomes eq.4d) 14×B = 21×D – 7×A
Hint #11
Substitute 5×D for 7×A in eq.4d: 14×B = 21×D – 5×D which becomes 14×B = 16×D Divide by 16: 14×B ÷ 16 = 16×D ÷ 16 which makes ⅞×B = D and also makes E = D = ⅞×B
Hint #12
Substitute (⅞×B) for D in eq.4c: 2×B = 3×(⅞×B) – A which becomes 2×B = 2⅝×B – A In the above equation, add A to both sides, and subtract 2×B from both sides: 2×B + A – 2×B = 2⅝×B – A + A – 2×B which makes A = ⅝×B
Solution
Substitute ⅝×B for A, B for C and F, and ⅞×B for D and E in eq.1: ⅝×B + B + B + ⅞×B + ⅞×B + B = 43 which simplifies to 5⅜×B = 43 Divide both sides of the equation above by 5⅜: 5⅜×B ÷ 5⅜ = 43 ÷ 5⅜ which means B = 8 making A = ⅝×B = ⅝ × 8 = 5 C = F = B = 8 D = E = ⅞×B = ⅞ × 8 = 7 and ABCDEF = 588778