Puzzle for August 21, 2022 ( )
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.
Scratchpad
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Hint #1
Add B to both sides of eq.2: B + C + B = E – B + B which becomes eq.2a) 2×B + C = E Add E to both sides of eq.5: C – E + E = D – B – C + E which becomes C = D – B – C + E which may be written as eq.5a) C = D + E – B – C
Hint #2
In eq.5a, replace E with 2×B + C (from eq.2a): C = D + 2×B + C – B – C which becomes eq.5b) C = D + B
Hint #3
Add A and D to both sides of eq.4: E – A + A + D = A + C – D + A + D which becomes E + D = 2×A + C which may be written as eq.4a) D + E = 2×A + C
Hint #4
In eq.5a, replace D + E with 2×A + C (from eq.4a): C = 2×A + C – B – C which becomes C = 2×A – B Add B to both sides of the equation above: C + B = 2×A – B + B which becomes eq.5c) C + B = 2×A
Hint #5
Add F, A, and D to both sides of eq.3: A – F + F + A + D = C – A – D + F + A + D which becomes eq.3a) 2×A + D = C + F
Hint #6
In eq.3a, substitute C + B for 2×A (from eq.5c): C + B + D = C + F Subtract C from each side of the above equation: C + B + D – C = C + F – C which becomes B + D = F which may be written as eq.3b) D + B = F
Hint #7
Substitute F for D + B (from eq.3b) into eq.5b: C = F
Hint #8
Subtract the left and right sides of eq.4 from the left and right sides of eq.3, respectively: A – F – (E – A) = C – A – D – (A + C – D) which becomes A – F – E + A = C – A – D – A – C + D which becomes 2×A – F – E = –2×A Add F, E, and 2×A to both sides of the above equation: 2×A – F – E + F + E + 2×A = –2×A + F + E + 2×A which simplifies to eq.3c) 4×A = E + F
Hint #9
Substitute 4×A for E + F (from eq.3c) into eq.6: A + D = 4×A – D In the equation above, subtract A from both sides, and add D to both sides: A + D – A + D = 4×A – D – A + D which becomes 2×D = 3×A Divide both sides by 2: 2×D ÷ 2 = 3×A ÷ 2 which makes D = 1½×A
Hint #10
Substitute 1½×A for D, and C for F in eq.3a: 2×A + 1½×A = C + C which becomes 3½×A = 2×C Divide both sides of the above equation by 2: 3½×A ÷ 2 = 2×C ÷ 2 which makes 1¾×A = C and also makes 1¾×A = C = F
Hint #11
Substitute 1¾×A for F in eq.3c: 4×A = E + 1¾×A Subtract 1¾×A from each side of the equation above: 4×A – 1¾×A = E + 1¾×A – 1¾×A which makes 2¼×A = E
Hint #12
Substitute 1¾×A for C in eq.5c: 1¾×A + B = 2×A Subtract 1¾×A from each side of the equation above: 1¾×A + B – 1¾×A = 2×A – 1¾×A which makes B = ¼×A
Solution
Substitute ¼×A for B, 1¾×A for C and F, 1½×A for D, and 2¼×A for E in eq.1: A + ¼×A + 1¾×A + 1½×A + 2¼×A + 1¾×A = 34 which simplifies to 8½×A = 34 Divide both sides of the above equation by 8½: 8½×A ÷ 8½ = 34 ÷ 8½ which means A = 4 making B = ¼×A = ¼ × 4 = 1 C = F = 1¾×A = 1¾ × 4 = 7 D = 1½×A = 1½ × 4 = 6 E = 2¼×A = 2¼ × 4 = 9 and ABCDEF = 417697