Puzzle for July 12, 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.
* BC and CD are 2-digit numbers (not B×C or C×D).
Scratchpad
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Hint #1
In eq.5, replace E with C + D (from eq.3): CD = C + C + D which may be written as 10×C + D = 2×C + D Subtract both 2×C and D from each side of the above equation: 10×C + D – 2×C – D = 2×C + D – 2×C – D which simplifies to 8×C = 0 which means C = 0
Hint #2
In eq.3, replace C with 0: E = 0 + D which means E = D
Hint #3
In eq.4, substitute D for E, and 0 for C: A – D + F = B + 0 + D In the above equation, add D to both sides, and subtract A from each side: A – D + F + D – A = B + 0 + D + D – A which becomes eq.4a) F = B + 2×D – A
Hint #4
Substitute B + 2×D – A for F (from eq.4a) in eq.2: D + B + 2×D – A = A + B which becomes B + 3×D – A = A + B In the above equation, subtract B from both sides, and add A to both sides: B + 3×D – A – B + A = A + B – B + A which simplifies to 3×D = 2×A Divide both sides by 2: 3×D ÷ 2 = 2×A ÷ 2 which makes 1½×D = A
Hint #5
eq.6 may be written as: 10×B + C = B + (D×E) Substitute 0 for C, and D for E in in the equation above: 10×B + 0 = B + (D×D) Subtract B from both sides: 10×B + 0 – B = B + (D×D) – B which makes 9×B = D×D which may be written as 9×B = D²
Hint #6
Multiply both sides of eq.2 by 9: 9×(D + F) = 9×(A + B) which is the same as 9×D + 9×F = 9×A + 9×B Substitute (1½×D) for A, and D² for 9×B in the above equation: 9×D + 9×F = 9×(1½×D) + D² which becomes 9×D + 9×F = 13½×D + D² Subtract 9×D from each side: 9×D + 9×F – 9×D = 13½×D + D² – 9×D which becomes eq.2a) 9×F = 4½×D + D²
Hint #7
Multiply both sides of eq.1 by 9: 9×(A + B + C + D + E + F) = 9×32 which is equivalent to 9×A + 9×B + 9×C + 9×D + 9×E + 9×F = 288 Substitute (1½×D) for A, D² for 9×B, 0 for C, D for E, and 4½×D + D² for 9×F (from eq.2a) in the above equation: 9×(1½×D) + D² + 9×0 + 9×D + 9×D + 4½×D + D² = 288 which becomes 13½×D + 2×D² + 22½×D = 288 which becomes 2×D² + 36×D = 288 Subtract 288 from each side: 2×D² + 36×D – 288 = 288 – 288 which becomes eq.1a) 2×D² + 36×D – 288 = 0
Solution
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.1a yields: D = {(–1)×36 ± sq.rt.[36² – (4 × 2 × (–288))]} ÷ (2 × 2) which becomes D = {–36 ± sq.rt.[1296 – (–2304)]} ÷ 4 which becomes D = {–36 ± sq.rt.[3600]} ÷ 4 which is equivalent to D = {–36 ± 60} ÷ 4 In the above equation, either: D = {–36 + 60} ÷ 4 = 24 ÷ 4 = 6 or: D = {–36 – 60} ÷ 4 = –96 ÷ 4 = –24 Since D must be non-negative, then D ≠ –24 and therefore D = 6 making A = 1½×D = 1½ × 6 = 9 9×B = D² = 6² = 36 making 9×B ÷ 9 = 36 ÷ 9 which means B = 4 E = D = 6 9×F = 4½×D + D² = 4½×6 + 6² = 27 + 36 = 63 making 9×F ÷ 9 = 63 ÷ 9 which means F = 7 (from eq.2a) and ABCDEF = 940667