Puzzle for April 2, 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 non-negative integer.
* CD is a 2-digit number (not C×D).
Scratchpad
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Hint #1
Add CD to both sides of eq.5: E – C + CD = A – CD + F + CD which becomes E – C + CD = A + F which may be written as E – C + 10×C + D = A + F which becomes eq.5a) E + 9×C + D = A + F
Hint #2
eq.6 may be written as: 10×C + D – E + F = A + B + C which may be written as 10×C + D – E + F = A + B + C Subtract C from each side of the above equation: 10×C + D – E + F – C = A + B + C – C which becomes eq.6a) 9×C + D – E + F = A + B
Hint #3
Subtract the left and right sides of eq.6a from the left and right sides of eq.5a, respectively: E + 9×C + D – (9×C + D – E + F) = A + F – (A + B) which is equivalent to E + 9×C + D – 9×C – D + E – F = A + F – A – B which becomes 2×E – F = F – B Add F and B to both sides of the above equation: 2×E – F + F + B = F – B + F + B which becomes eq.5b) 2×E + B = 2×F
Hint #4
Add D and F to both sides of eq.3: D + F + D + F = A + B – D + E – F + D + F which becomes 2×D + 2×F = A + B + E In the above equation, replace 2×F with 2×E + B (from eq.5b): 2×D + 2×E + B = A + B + E Subtract B and E from each side: 2×D + 2×E + B – B – E = A + B + E – B – E which becoems eq.3a) 2×D + E = A
Hint #5
In eq.3a, replace E with A + D (from eq.2): 2×D + A + D = A which becomes 3×D + A = A Subtract A from each side of the above equation: 3×D + A – A = A – A which makes 3×D = 0 which means D = 0
Hint #6
In eq.2, replace D with 0: E = A + 0 which makes E = A
Hint #7
In eq.5a, substitute A for E, and 0 for D: A + 9×C + 0 = A + F which becomes A + 9×C = A + F Subtract A from both sides of the equation above: A + 9×C – A = A + F – A which makes 9×C = F
Hint #8
eq.4 may be written as: A – C = (D + E + F) ÷ 3 Multiply both sides of the equation above by 3: 3 × (A – C) = 3 × (D + E + F) ÷ 3 which becomes eq.4a) 3×A – 3×C = D + E + F
Hint #9
In eq.4a, substitute 0 for D, A for E, and 9×C for F: 3×A – 3×C = 0 + A + 9×C In the above equation, subtract A from each side, and add 3×C to both sides: 3×A – 3×C – A + 3×C = 0 + A + 9×C – A + 3×C which becomes 2×A = 12×C Divide both sides by 2: 2×A ÷ 2 = 12×C ÷ 2 which makes A = 6×C and also makes E = A = 6×C
Hint #10
Substitute (6×C) for E, and (9×C) for F in eq.5b: 2×(6×C) + B = 2×(9×C) which becomes 12×C + B = 18×C Subtract 12×C from each side of the equation above: 12×C + B – 12×C = 18×C – 12×C which makes B = 6×C
Solution
Substitute 6×C for A and B and E, 0 for D, and 9×C for F in eq.1: 6×C + 6×C + C + 0 + 6×C + 9×C = 28 which simplifies to 28×C = 28 Divide both sides of the equation above by 28: 28×C ÷ 28 = 28 ÷ 28 which means C = 1 making A = B = E = 6×C = 6 × 1 = 6 F = 9×C = 9 × 1 = 9 and ABCDEF = 661069