Puzzle for October 19, 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.
* AB is a 2-digit number (not A×B). ABC and DEF are 3-digit numbers (not A×B×C or D×E×F).
Scratchpad
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Hint #1
In eq.4, replace B with C – D (from eq.5): A + C = C – D + E Subtract C from both sides of the above equation: A + C – C = C – D + E – C which becomes eq.4a) A = –D + E
Hint #2
In eq.2, replace A with –D + E (from eq.4a): E + F = –D + E + D which becomes E + F = E Subtract E from both sides of the equation above: E + F – E = E – E which makes F = 0
Hint #3
eq.2 may be written as: A + D = E + F Add the left and right sides of the above equation to the left and right sides of eq.3, respectively: B – A + A + D = E – F + E + F which becomes eq.3a) B + D = 2×E
Hint #4
Add D to both sides of eq.5: C – D + D = B + D which becomes C = B + D In the above equation, substitute 2×E for B + D (from eq.3a): eq.5a) C = 2×E
Hint #5
eq.6 may be written as: 100×D + 10×E + F + 10×A + B = 100×A + 10×B + C – D In the equation above, subtract 10×A and B from both sides, and add D to both sides: 100×D + 10×E + F + 10×A + B – 10×A – B + D = 100×A + 10×B + C – D – 10×A – B + D which becomes 101×D + 10×E + F = 90×A + 9×B + C Substitute 0 for F, and 2×E for C (from eq.5a): 101×D + 10×E + 0 = 90×A + 9×B + 2×E Subtract 2×E from both sides: 101×D + 10×E + 0 – 2×E = 90×A + 9×B + 2×E – 2×E which becomes eq.6a) 101×D + 8×E = 90×A + 9×B
Hint #6
Substitute 2×E for C (from eq.5a) in eq.5: 2×E – D = B Substitute (–D + E) for A (from eq.4a), and (2×E – D) for B in eq.6a: 101×D + 8×E = 90×(–D + E) + 9×(2×E – D) which is equivalent to 101×D + 8×E = –90×D + 90×E + 18×E – 9×D which becomes 101×D + 8×E = –99×D + 108×E In the above equation, add 99×D to both sides, and subtract 8×E from both sides: 101×D + 8×E + 99×D – 8×E = –99×D + 108×E + 99×D – 8×E which means 200×D = 100×E Divide both sides by 100: 200×D ÷ 100 = 100×E ÷ 100 which makes 2×D = E
Hint #7
Substitute 2×D for E in eq.4a: A = –D + 2×D which makes A = D
Hint #8
Substitute (2×D) for E in eq.5a: C = 2×(2×D) which means C = 4×D
Hint #9
Substitute 4×D for C in eq.5: 4×D – D = B which makes 3×D = B
Solution
Substitute D for A, 3×D for B, 4×D for C, 2×D for E, and 0 for F in eq.1: D + 3×D + 4×D + D + 2×D + 0 = 11 which simplifies to 11×D = 11 Divide both sides of the equation above by 11: 11×D ÷ 11 = 11 ÷ 11 which means D = 1 making A = D = 1 B = 3×D = 3 × 1 = 3 C = 4×D = 4 × 1 = 4 E = 2×D = 2 × 1 = 2 and ABCDEF = 134120