Puzzle for March 27, 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.
* BC and EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
eq.5 may be written as: 10×B + C – D = 10×E + F Add D to both sides of the above equation: 10×B + C – D + D = 10×E + F + D which becomes eq.5a) 10×B + C = 10×E + F + D which may be written as eq.5b) 9×B + B + C = 10×E + F + D
Hint #2
Add C and F to both sides of eq.2: D + E – C + C + F = B – F + C + F which becomes eq.2a) D + E + F = B + C
Hint #3
In eq.5b, replace B + C with D + E + F (from eq.2a): 9×B + D + E + F = 10×E + F + D Subtract D, E, and F from each side of the above equation: 9×B + D + E + F – D – E – F = 10×E + F + D – D – E – F which simplifies to 9×B = 9×E Divide both sides by 9: 9×B ÷ 9 = 9×E ÷ 9 which makes B = E
Hint #4
In eq.5a, replace E with B: 10×B + C = 10×B + F + D Subtract 10×B from each side of the equation above: 10×B + C – 10×B = 10×B + F + D – 10×B which becomes C = F + D which is the same as eq.5c) C = D + F
Hint #5
eq.6 may be written as: E = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × (C + D + F) ÷ 3 which becomes eq.6a) 3×E = C + D + F
Hint #6
In eq.6a, substitute C for D + F (from eq.5c): 3×E = C + C which makes 3×E = 2×C Divide both sides of the above equation by 2: 3×E ÷ 2 = 2×C ÷ 2 which makes 1½×E = C
Hint #7
In eq.4, substitute E for B: A + C + D – E – F = E + E + F which becomes A + C + D – E – F = 2×E + F Add E and F to both sides of the above equation: A + C + D – E – F + E + F = 2×E + F + E + F which becomes eq.4a) A + C + D = 3×E + 2×F
Hint #8
Substitute C + D + F for 3×E (from eq.6a) into eq.4a: A + C + D = C + D + F + 2×F which becomes A + C + D = C + D + 3×F Subtract C and D from each side of the equation above: A + C + D – C – D = C + D + 3×F – C – D which simplifies to eq.4b) A = 3×F
Hint #9
Substitute B for E, and 3×F for A (from eq.4b) in eq.3: C + D + B – F = 3×F + B – D In the equation above, subtract B from both sides, and add F and D to both sides: C + D + B – F – B + F + D = 3×F + B – D – B + F + D which simplifies to eq.3a) C + 2×D = 4×F
Hint #10
Substitute D + F for C (from eq.5c) into eq.3a: D + F + 2×D = 4×F which becomes 3×D + F = 4×F Subtract F from each side of the above equation: 3×D + F – F = 4×F – F which makes 3×D = 3×F Divide both sides by 3: 3×D ÷ 3 = 3×F ÷ 3 which makes D = F
Hint #11
Substitute 1½×E for C, and F for D in eq.5c: 1½×E = F + F which becomes 1½×E = 2×F Divide both sides by 2: 1½×E ÷ 2 = 2×F ÷ 2 which makes ¾×E = F and also makes ¾×E = F = D
Hint #12
Substitute (¾×E) for F in eq.4b: A = 3×(¾×E) which makes A = 2¼×E
Solution
Substitute 2¼×E for A, E for B, 1½×E for C, and ¾×E for D and F in eq.1: 2¼×E + E + 1½×E + ¾×E + E + ¾×E = 29 which simplifies to 7¼×E = 29 Divide both sides of the above equation by 7¼: 7¼×E ÷ 7¼ = 29 ÷ 7¼ which means E = 4 making A = 2¼×E = 2¼ × 4 = 9 B = E = 4 C = 1½×E = 1½ × 4 = 6 D = F = ¾×E = ¾ × 4 = 3 and ABCDEF = 946343