Puzzle for August 14, 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 positive integer.
* AB, CD, and DE are 2-digit numbers (not A×B, C×D, or D×E).
Scratchpad
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Hint #1
Subtract the left and right sides of eq.4 from the left and right sides of eq.3, respectively: C + E – (B + C) = A + F – (E + F) which becomes C + E – B – C = A + F – E – F which becomes E – B = A – E Add B and E to both sides of the equation above: E – B + B + E = A – E + B + E which becomes eq.3a) 2×E = A + B
Hint #2
Add the left and right sides of eq.2 to the left and right sides of eq.1, respectively: A + E + B + E = B + F + A + C which becomes A + 2×E + B = B + F + A + C Subtract A and B from both sides of the equation above: A + 2×E + B – A – B = B + F + A + C – A – B which becomes eq.1a) 2×E = F + C
Hint #3
eq.5 may be written as: 10×A + B – (10×C + D) = F – D which becomes 10×A + B – 10×C – D = F – D Add 10×C and D to both sides of the above equation: 10×A + B – 10×C – D + 10×C + D = F – D + 10×C + D which becomes 10×A + B = F + 10×C which may be written as eq.5a) 9×A + A + B = F + C + 9×C
Hint #4
In eq.5a, replace A + B with 2×E (from eq.3a), and F + C with 2×E (from eq.1a): 9×A + 2×E = 2×E + 9×C Subtract 2×E from each side of the above equation: 9×A + 2×E – 2×E = 2×E + 9×C – 2×E which becomes 9×A = 9×C Divide both sides by 9: 9×A ÷ 9 = 9×C ÷ 9 which makes A = C
Hint #5
In eq.3, replace C with A: A + E = A + F Subtract A from each side of the equation above: A + E – A = A + F – A which makes E = F
Hint #6
In eq.1, replace F with E: A + E = B + E Subtract E from each side of the equation above: A + E – E = B + E – E which makes A = B and also makes C = A = B
Hint #7
In eq.2, substitute A for B: A + E = A + C Subtract A from each side of the above equation: A + E – A = A + C – A which makes E = C and therefore makes A = B = C = E = F
Hint #8
eq.6 may be written as: 10×D + E – B = A + (C ÷ E) + (F ÷ C) In the equation above, substitute A for B, C, E, and F: 10×D + A – A = A + (A ÷ A) + (A ÷ A) which becomes 10×D = A + 1 + 1 which makes eq.6a) 10×D = A + 2
Hint #9
Since A is a one-digit positive integer, then A ≤ 9 Adding 2 to both sides of the above inequality yields: A + 2 ≤ 9 + 2 which means A + 2 ≤ 11 Combining the above inequality with eq.6a yields: 10×D = A + 2 ≤ 11 which means ie.6b) 10×D ≤ 11
Hint #10
To make inequality ie.6b true, check several possible values for D: If D = 1 then 10×D = 10×1 = 10 ≤ 11 If D = 2 then 10×D = 10×2 = 20 > 11 If D > 2 then 10×D > 11 The only one-digit positive integer for D that makes inequality ie.6b true is: D = 1
Solution
Substitute 1 for D in eq.6a: 10×1 = A + 2 which becomes 10 = A + 2 Subtract 2 from both sides of the equation above: 10 – 2 = A + 2 – 2 which makes 8 = A and also makes A = B = C = E = F = 8 and ABCDEF = 888188