Puzzle for September 19, 2019 ( )
Scratchpad
Find the 5-digit number ABCDE by solving the following equations:
A, B, C, D, and E each represent a one-digit non-negative integer.
* AB, BC, and CD are 2-digit numbers (not A×B, B×C, or C×D).
Scratchpad
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Hint #1
eq.1 may be written as: A + C + B + E + D = 24 In the equation above, substitute D for A + C (from eq.2) and for B + E (from eq.4): D + D + D = 24 which means 3×D = 24 Divide both sides by 3: 3×D ÷ 3 = 24 ÷ 3 which makes D = 8
Hint #2
In eq.4, replace D with 8: B + E = 8 Subtract B from both sides of the equation above: B + E – B = 8 – B which becomes eq.4a) E = 8 – B
Hint #3
In eq.3, substitute 8 – B for E (from eq.4a): 8 – B – B = A + B Subtract B from both sides of the above equation: 8 – B – B – B = A + B – B which becomes eq.3a) 8 – 3×B = A
Hint #4
Substitute 8 for D, and 8 – 3×B for A (from eq.3a) in eq.2: 8 = 8 – 3×B + C In the equation above, add 3×B to both sides, and subtract 8 from each side: 8 + 3×B – 8 = 8 – 3×B + C + 3×B – 8 which simplifies to 3×B = C
Solution
eq.5 may be written as: 10×A + B = 10×B + C + 10×C + D Subtract B from each side of the above equation: 10×A + B – B = 10×B + C + 10×C + D – B which becomes 10×A = 9×B + 11×C + D Substitute (8 – 3×B) for A (from eq.3a), (3×B) for C, and 8 for D: 10×(8 – 3×B) = 9×B + 11×(3×B) + 8 which becomes 80 – 30×B = 9×B + 33×B + 8 which becomes 80 – 30×B = 42×B + 8 Add 30×B to both sides, and subtract 8 from both sides: 80 – 30×B + 30×B – 8 = 42×B + 8 + 30×B – 8 which simplifies to 72 = 72×B Divide both sides by 72: 72 ÷ 72 = 72×B ÷ 72 which means 1 = B making A = 8 – 3×B = 8 – 3×1 = 5 (from eq.3a) C = 3×B = 3×1 = 3 E = 8 – B = 8 – 1 = 7 (from eq.4a) and ABCDE = 51387