Puzzle for September 9, 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.
Scratchpad
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Hint #1
In eq.3, add C to both sides, and subtract E from both sides: C – F + C – E = B – C + E + C – E which becomes eq.3a) 2×C – F – E = B Add D, F, and A to both sides of eq.4: A – D – F + D + F + A = B + D + F – A – E + D + F + A which becomes eq.4a) 2×A = B + 2×D + 2×F – E
Hint #2
In eq.4a, replace B with 2×C – F – E (from eq.3a): 2×A = 2×C – F – E + 2×D + 2×F – E which becomes 2×A = 2×C + 2×D – 2×E + F which may be written as eq.4b) 2×A = 2×(C + D) – 2×E + F Add E to both sides of eq.2: C + D – E + E = A + E + E which becomes eq.2a) C + D = A + 2×E
Hint #3
In eq.4b, replace C + D with A + 2×E (from eq.2a): 2×A = 2×(A + 2×E) – 2×E + F which becomes 2×A = 2×A + 4×E – 2×E + F which becomes 2×A = 2×A + 2×E + F Subtract 2×A from both sides of the above equation: 2×A – 2×A = 2×A + 2×E + F – 2×A which becomes 0 = 2×E + F Since E and F are non-negative integers, the only value for E and F that makes the above equation true is: E = 0 and F = 0
Hint #4
In eq.3a, substitute 0 for E and F: 2×C – 0 – 0 = B which makes 2×C = B
Hint #5
eq.5 may be written as: D = (B + C + E + F) ÷ 4 Multiply both sides of the above equation by 4: D × 4 = ((B + C + E + F) ÷ 4) × 4 which becomes eq.5a) 4×D = B + C + E + F
Hint #6
Substitute 2×C for B, and 0 for E and F in eq.5a: 4×D = 2×C + C + 0 + 0 which becomes 4×D = 3×C Divide both sides of the above equation by 4: 4×D ÷ 4 = 3×C ÷ 4 which makes D = ¾×C
Hint #7
Substitute ¾×C for D, and 0 for E in eq.2a: C + ¾×C = A + 2×0 which makes 1¾×C = A
Solution
Substitute 1¾×C for A, 2×C for B, ¾×C for D, and 0 for E and F in eq.1: 1¾×C + 2×C + C + ¾×C + 0 + 0 = 22 which simplifies to 5½×C = 22 Divide both sides of the above equation by 5½: 5½×C ÷ 5½ = 22 ÷ 5½ which means C = 4 making A = 1¾×C = 1¾ × 4 = 7 B = 2×C = 2 × 4 = 8 D = ¾×C = ¾ × 4 = 3 and ABCDEF = 784300