Puzzle for June 13, 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, CD, and EF are 2-digit numbers (not B×C, C×D, or E×F).
Scratchpad
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Hint #1
In eq.2, replace C + D with B + F (from eq.3): A + B = B + F + E Subtract B from each side of the above equation: A + B – B = B + F + E – B which becomes eq.2a) A = F + E
Hint #2
Subtract both A and D from each side of eq.5: D – E + F – A – D = A + C – A – D which becomes –E + F – A = C – D In the above equation, replace C – D with A – F (from eq.4): –E + F – A = A – F Add A, E, and F to both sides: –E + F – A + A + E + F = A – F + A + E + F which simplifies to eq.5a) 2×F = 2×A + E
Hint #3
In eq.5a, substitute (F + E) for A (from eq.2a): 2×F = 2×(F + E) + E which is equivalent to 2×F = 2×F + 2×E + E which becomes 2×F = 2×F + 3×E Subtract 2×F from both sides of the above equation: 2×F – 2×F = 2×F + 3×E – 2×F which becomes 0 = 3×E which means 0 = E
Hint #4
Substitute 0 for E in eq.2a: A = F + 0 which makes A = F
Hint #5
Substitute A for F in eq.4: C – D = A – A which means C – D = 0 Add D to both sides: C – D + D = 0 + D which makes C = D
Hint #6
Substitute C for D, and 0 for E in eq.2: A + B = C + C + 0 which becomes eq.2b) A + B = 2×C
Hint #7
eq.6 may be written as: 10×B + C = A + 10×C + D + 10×E + F Subtract C from both sides of the equation above: 10×B + C – C = A + 10×C + D + 10×E + F – C which becomes 10×B = A + 9×C + D + 10×E + F Substitute C for D, 0 for E, and A for F: 10×B = A + 9×C + C + 10×0 + A which becomes 10×B = 2×A + 10×C which may be written as eq.6a) 10×B = 2×A + 5×(2×C)
Hint #8
Substitute A + B for 2×C (from eq.2b) in eq.6a: 10×B = 2×A + 5×(A + B) which is equivalent to 10×B = 2×A + 5×A + 5×B which becomes 10×B = 7×A + 5×B Subtract 5×B from each side of the equation above: 10×B – 5×B = 7×A + 5×B – 5×B which becomes 5×B = 7×A Divide both sides by 5: 5×B ÷ 5 = 7×A ÷ 5 which makes B = 1⅖×A
Hint #9
Substitute 1⅖×A for B in eq.2b: A + 1⅖×A = 2×C which becomes 2⅖×A = 2×C Divide both sides of the above equation by 2: 2⅖×A ÷ 2 = 2×C ÷ 2 which makes 1⅕×A = C and also makes D = C = 1⅕×A
Solution
Substitute 1⅖×A for B, 1⅕×A for C and D, 0 for E, and A for F in eq.1: A + 1⅖×A + 1⅕×A + 1⅕×A + 0 + A = 29 which simplifies to 5⅘×A = 29 Divide both sides of the equation above by 5⅘: 5⅘×A ÷ 5⅘ = 29 ÷ 5⅘ which means A = 5 making B = 1⅖×A = 1⅖ × 5 = 7 C = D = 1⅕×A = 1⅕ × 5 = 6 F = A = 5 and ABCDEF = 576605