Puzzle for May 1, 2021 ( )
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, CD, DE, and EF are 2-digit numbers (not A×B, C×D, D×E, or E×F).
** ABC and DEF are 3-digit numbers (not A×B×C or D×E×F).
Scratchpad
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Hint #1
eq.6 may be written as: 100×D + 10×E + F = 100×A + 10×B + C + 10×C + D which becomes 100×D + 10×E + F = 100×A + 10×B + 11×C + D Subtract D from both sides of the above equation: 100×D + 10×E + F – D = 100×A + 10×B + 11×C + D – D which becomes eq.6a) 99×D + 10×E + F = 100×A + 10×B + 11×C
Hint #2
eq.4 may be written as: 10×C + D = B – C + 10×E + F In the equation above, add C to both sides, and subtract B from both sides: 10×C + D + C – B = B – C + 10×E + F + C – B which becomes eq.4a) 11×C + D – B = 10×E + F
Hint #3
In eq.6a, replace 10×E + F with 11×C + D – B (from eq.4a): 99×D + 11×C + D – B = 100×A + 10×B + 11×C which becomes 100×D + 11×C – B = 100×A + 10×B + 11×C In the above equation, subtract 11×C from both sides, and add B to both sides: 100×D + 11×C – B – 11×C + B = 100×A + 10×B + 11×C – 11×C + B which becomes eq.6b) 100×D = 100×A + 11×B
Hint #4
In eq.2, subtract D from both sides, and add B to both sides: D – D + B = A – B – D + B which becomes eq.2a) B = A – D Substitute (A – D) for B (from eq.2a) in eq.6b: 100×D = 100×A + 11×(A – D) which is equivalent to 100×D = 100×A + 11×A – 11×D which becomes 100×D = 111×A – 11×D Add 11×D to both sides of the above equation: 100×D + 11×D = 111×A – 11×D + 11×D which becomes 111×D = 111×A Divide both sides of the above equation by 111: 111×D ÷ 111 = 111×A ÷ 111 which makes D = A
Hint #5
Substitute A for D in eq.2a: B = A – A which means B = 0
Hint #6
Substitute 0 for B, and A for D in eq.3: 0 + C – A – A = A + A which becomes C – 2×A = 2×A Add 2×A to both sides of the equation above: C – 2×A + 2×A = 2×A + 2×A which makes C = 4×A
Hint #7
Substitute (4×A) for C, A for D, and 0 for B in eq.4a: 11×(4×A) + A – 0 = 10×E + F which becomes 44×A + A = 10×E + F which becomes eq.4b) 45×A = 10×E + F
Hint #8
eq.5 may be written as: 10×A + B + C = 10×D + E – (A ÷ D) Substitute 0 for B, 4×A for C, and A for D in the above equation: 10×A + 0 + 4×A = 10×A + E – (A ÷ A) which becomes 14×A = 10×A + E – 1 In the equation above, subtract 10×A from both sides, and add 1 to both sides: 14×A – 10×A + 1 = 10×A + E – 1 – 10×A + 1 which becomes eq.5a) 4×A + 1 = E
Hint #9
Substitute (4×A + 1) for E (from eq.5a) into eq.4b: 45×A = 10×(4×A + 1) + F which becomes 45×A = 40×A + 10 + F Subtract 40×A and 10 from both sides of the equation above: 45×A – 40×A – 10 = 40×A + 10 + F – 40×A – 10 which becomes eq.4c) 5×A – 10 = F
Solution
Substitute 0 for B, 4×A for C, A for D, 4×A + 1 for E (from eq.5a), and 5×A – 10 for F (from eq.4c) in eq.1: A + 0 + 4×A + A + 4×A + 1 + 5×A – 10 = 21 which simplifies to 15×A – 9 = 21 Add 9 to both sides of the above equation: 15×A – 9 + 9 = 21 + 9 which makes 15×A = 30 Divide both sides by 15: 15×A ÷ 15 = 30 ÷ 15 which means A = 2 making C = 4×A = 4×2 = 8 D = A = 2 E = 4×A + 1 = 4×2 + 1 = 8 + 1 = 9 (from eq.5a) F = 5×A – 10 = 5×2 – 10 = 10 – 10 = 0 (from eq.4c) and ABCDEF = 208290