Puzzle for August 15, 2020 ( )
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.
* BC, CD, and EF are 2-digit numbers (not B×C, C×D, or E×F).
Scratchpad
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Hint #1
In eq.2, replace B with C + D (from eq.3): D + E + F = C + D + C Subtract D from each side of the above equation: D + E + F – D = C + D + C – D which simplifies to eq.2a) E + F = 2×C Subtract E from both sides of eq.2a: E + F – E = 2×C – E which becomes eq.2b) F = 2×C – E
Hint #2
eq.6 may be written as: 10×B + C = 10×C + D + 10×E + F – C which becomes 10×B + C = 9×C + D + 10×E + F Subtract C from each side of the equation above: 10×B + C – C = 9×C + D + 10×E + F – C which becomes eq.6a) 10×B = 8×C + D + 10×E + F
Hint #3
In eq.6a, substitute (C + D) for B (from eq.3): 10×(C + D) = 8×C + D + 10×E + F which becomes 10×C + 10×D = 8×C + D + 10×E + F Subtract 8×C and D from both sides of the above equation: 10×C + 10×D – 8×C – D = 8×C + D + 10×E + F – 8×C – D which simplifies to eq.6b) 2×C + 9×D = 10×E + F
Hint #4
In eq.6b, substitute E + F for 2×C (from eq.2a): E + F + 9×D = 10×E + F Subtract both E and F from each side of the equation above: E + F + 9×D – E – F = 10×E + F – E – F which simplifies to 9×D = 9×E Divide both sides by 9: 9×D ÷ 9 = 9×E ÷ 9 which makes D = E
Hint #5
Subtract D from both sides of eq.4: A + D – D = B + F – D which becomes A = B + F – D Substitute D for E, and B + F – D for A in eq.5: B + D = B + F – D + C + F which becomes B + D = B – D + C + 2×F In the equation above, add D to both sides, and subtract B from both sides: B + D + D – B = B – D + C + 2×F + D – B which becomes eq.5a) 2×D = C + 2×F
Hint #6
Substitute E for D, and (2×C – E) for F (from eq.2b) in eq.5a: 2×E = C + 2×(2×C – E) which becomes 2×E = C + 4×C – 2×E which becomes 2×E = 5×C – 2×E Add 2×E to both sides of the above equation: 2×E + 2×E = 5×C – 2×E + 2×E which becomes 4×E = 5×C Divide both sides by 4: 4×E ÷ 4 = 5×C ÷ 4 which makes E = 1¼×C and also makes D = E = 1¼×C
Hint #7
Substitute 1¼×C for D in eq.3: C + 1¼×C = B which makes 2¼×C = B
Hint #8
Substitute 1¼×C for E in eq.2b: F = 2×C – 1¼×C which makes F = ¾×C
Hint #9
Substitute 1¼×C for D, 2¼×C for B, and ¾×C for F in eq.4: A + 1¼×C = 2¼×C + ¾×C which becomes A + 1¼×C = 3×C Subtract 1¼×C from both sides of the equation above: A + 1¼×C – 1¼×C = 3×C – 1¼×C which makes A = 1¾×C
Solution
Substitute 1¾×C for A, 2¼×C for B, 1¼×C for D and E, and ¾×C for F in eq.1: 1¾×C + 2¼×C + C + 1¼×C + 1¼×C + ¾×C = 33 which becomes 8¼×C = 33 Divide both sides of the above equation by 8¼: 8¼×C ÷ 8¼ = 33 ÷ 8¼ which means C = 4 making A = 1¾×C = 1¾ × 4 = 7 B = 2¼×C = 2¼ × 4 = 9 D = E = 1¼×C = 1¼ × 4 = 5 F = ¾×C = ¾ × 4 = 3 and ABCDEF = 794553