Puzzle for February 8, 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 DE are 2-digit numbers (not B×C or D×E).
Scratchpad
Help Area
Hint #1
eq.6 may be written as: 10×B + C – (10×D + E) = A + B which is the same as 10×B + C – 10×D – E = A + B Subtract B from both sides of the above equation: 10×B + C – 10×D – E – B = A + B – B which becomes eq.6a) 9×B + C – 10×D – E = A
Hint #2
Subtract D from both sides of eq.3: A + D – D = C + E – D which becomes A = C + E – D In eq.6a, replace A with C + E – D: 9×B + C – 10×D – E = C + E – D In the above equation, add D and E to both sides, and subtract C from each side: 9×B + C – 10×D – E + D + E – C = C + E – D + D + E – C which simplifies to eq.6b) 9×B – 9×D = 2×E
Hint #3
Add B to both sides of eq.5: E – B + B = B – D + B which becomes eq.5a) E = 2×B – D In eq.6b, replace (2×B – D) for E (from eq.5a): 9×B – 9×D = 2×(2×B – D) which becomes 9×B – 9×D = 4×B – 2×D In the above equation, add 9×D to both sides, and subtract 4×B from each side: 9×B – 9×D + 9×D – 4×B = 4×B – 2×D + 9×D – 4×B which makes 5×B = 7×D Divide both sides by 5: 5×B ÷ 5 = 7×D ÷ 5 which makes B = 1⅖×D
Hint #4
Substitute (1⅖×D) for B in eq.5a: E = 2×(1⅖×D) – D which becomes E = 2⅘×D – D which makes E = 1⅘×D
Hint #5
Subtract D and C from both sides of eq.3: A + D – D – C = C + E – D – C which becomes eq.3a) A – C = E – D Subtract C and F from both sides of eq.4: C + D – C – F = A + F – C – F which becomes eq.4a) D – F = A – C
Hint #6
In eq.3a, substitute D – F for A – C (from eq.4a), and 1⅘×D for E: D – F = 1⅘×D – D which becomes D – F = ⅘×D In the above equation, add F to both sides, and subtract ⅘×D from both sides: D – F + F – ⅘×D = ⅘×D + F – ⅘×D which makes ⅕×D = F
Hint #7
Substitute 1⅖×D for B, ⅕×D for F, and 1⅘×D for E in eq.2: 1⅖×D + D + ⅕×D = A + 1⅘×D which becomes 2⅗×D = A + 1⅘×D Subtract 1⅘×D from each side: 2⅗×D – 1⅘×D = A + 1⅘×D – 1⅘×D which makes ⅘×D = A
Hint #8
Substitute ⅘×D for A, and ⅕×D for F in eq.4: C + D = ⅘×D + ⅕×D which becomes C + D = D Subtract D from both sides of the equation above: C + D – D = D – D which means C = 0
Solution
Substitute ⅘×D for A, 1⅖×D for B, 0 for C, 1⅘×D for E, and ⅕×D for F in eq.1: ⅘×D + 1⅖×D + 0 + D + 1⅘×D + ⅕×D = 26 which simplifies to 5⅕×D = 26 Divide both sides of the above equation by 5⅕: 5⅕×D ÷ 5⅕ = 26 ÷ 5⅕ which means D = 5 making A = ⅘×D = ⅘ × 5 = 4 B = 1⅖×D = 1⅖ × 5 = 7 E = 1⅘×D = 1⅘ × 5 = 9 F = ⅕×D = ⅕ × 5 = 1 and ABCDEF = 470591