Puzzle for May 9, 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 positive integer.
* BC is a 2-digit number (not B×C).
Scratchpad
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Hint #1
eq.5 may be written as: C + E = D + A + F In the equation above, replace A + F with C (from eq.2): C + E = D + C Subtract C from each side of the equation above: C + E - C = D + C - C which makes E = D
Hint #2
Subtract the left and right sides of eq.5 from the left and right sides of eq.4, respectively: C + F - (C + E) = A + E - (A + D + F) which is equivalent to C + F - C - E = A + E - A - D - F which becomes F - E = E - D - F In the above equation, replace D with E: F - E = E - E - F which becomes F - E = -F Add both E and F to each side: F - E + E + F = -F + E + F which makes 2×F = E and which also makes D = E = 2×F
Hint #3
Add B to both sides of eq.3: D + B + B = C - B + B which becomes eq.3a) D + 2×B = C In eq.1, substitute D + 2×B for C (from eq.3a): E + F = B + D + 2×B which becomes eq.1a) E + F = 3×B + D
Hint #4
In eq.1a, substitute 2×F for both E and D: 2×F + F = 3×B + 2×F Subtract 2×F from each side: 2×F + F - 2×F = 3×B + 2×F - 2×F which makes F = 3×B Divide both sides by 3: F ÷ 3 = 3×B ÷ 3 which makes ⅓×F = B
Hint #5
Substitute 2×F for D, and (⅓×F) for B in eq.3a: 2×F + 2×(⅓×F) = C which becomes 2×F + ⅔×F = C which makes 2⅔×F = C
Hint #6
Substitute 2⅔×F for C in eq.2: A + F = 2⅔×F Subtract F from each side: A + F - F = 2⅔×F - F which makes A = 1⅔×F
Solution
eq.6 may be written as: 10×B + C = E × F Substitute (⅓×F) for B, 2⅔×F for C, and 2×F for E in the above equation: 10×(⅓×F) + 2⅔×F = 2×F × F which becomes 3⅓×F + 2⅔×F = 2×F² which makes 6×F = 2×F² Divide both sides by 2×F: 6×F ÷ 2×F = 2×F² ÷ 2×F which means 3 = F making B = ⅓×F = ⅓ × 3 = 1 C = 2⅔×F = 2⅔ × 3 = 8 D = E = 2×F = 2 × 3 = 6 A = 1⅔×F = 1⅔ × 3 = 5 and ABCDEF = 518663