Puzzle for March 27, 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.
* DE and CD are 2-digit numbers (not D×E or C×D).
Scratchpad
Help Area
Hint #1
eq.3 may be written as: A – F + E = B + D + F In the above equation, replace A – F with C + F (from eq.2): C + F + E = B + D + F Subtract F from both sides: C + F + E – F = B + D + F – F which becomes eq.3a) C + E = B + D
Hint #2
eq.6 may be written as: C = (B + D + E) ÷ 3 Multiply both sides of the above equation by 3: C × 3 = ((B + D + E) ÷ 3) × 3 which becomes eq.6a) 3×C = B + D + E
Hint #3
In eq.6a, replace B + D with C + E (from eq.3a): 3×C = C + E + E which becomes 3×C = C + 2×E Subtract C from each side of the above equation: 3×C – C = C + 2×E – C which becomes 2×C = 2×E Divide both sides by 2: 2×C ÷ 2 = 2×E ÷ 2 which makes C = E
Hint #4
In eq.3a, substitute C for E: C + C = B + D which becomes eq.3b) 2×C = B + D
Hint #5
Add B, C, D, and E to both sides of eq.4: B – C + D – E + B + C + D + E = A – B – D + E + B + C + D + E which simplifies to 2×B + 2×D = A + C + 2×E which may be written as eq.4a) 2×(B + D) = A + C + 2×E
Hint #6
Substitute 2×C for B + D (from eq.3b), and C for E in eq.4a: 2×(2×C) = A + C + 2×C which becomes 4×C = A + 3×C Subtract 3×C from both sides of the above equation: 4×C – 3×C = A + 3×C – 3×C which makes C = A
Hint #7
Substitute A for C in eq.2: A + F = A – F In the above equation, subtract A from both sides, and add F to both sides: A + F – A + F = A – F – A + F which makes 2×F = 0 which means F = 0
Hint #8
eq.5 may be written as: 10×D + E – B – C – D = B + 10×C + D which becomes 9×D + E – B – C = B + 10×C + D In the equation above, add B and C to both sides, and subtract D and E from both sides: 9×D + E – B – C + B + C – D – E = B + 10×C + D + B + C – D – E which simplifies to eq.5a) 8×D = 2×B + 11×C – E
Hint #9
Substitute C for E in eq.5a: 8×D = 2×B + 11×C – C which becomes 8×D = 2×B + 10×C Divide both sides of the above equation by 2: 8×D ÷ 2 = (2×B + 10×C) ÷ 2 which becomes 4×D = B + 5×C Subtract 5×C from both sides: 4×D – 5×C = B + 5×C – 5×C which becomes eq.5b) 4×D – 5×C = B
Hint #10
Substitute 4×D – 5×C for B (from eq.5b) into eq.3b: 2×C = 4×D – 5×C + D which becomes 2×C = 5×D – 5×C Add 5×C to both sides of the equation above: 2×C + 5×C = 5×D – 5×C + 5×C which makes 7×C = 5×D Divide both sides by 5: 7×C ÷ 5 = 5×D ÷ 5 which makes 1⅖×C = D
Hint #11
In eq.3b, substitute 1⅖×C for D: 2×C = B + 1⅖×C Subtract 1⅖×C from each side of the above equation: 2×C – 1⅖×C = B + 1⅖×C – 1⅖×C which makes ⅗×C = B
Solution
Substitute C for A and E, ⅗×C for B, 1⅖×C for D, and 0 for F in eq.1: C + ⅗×C + C + 1⅖×C + C + 0 = 25 which simplifies to 5×C = 25 Divide both sides of the above equation by 5: 5×C ÷ 5 = 25 ÷ 5 which means C = 5 making A = E = C = 5 B = ⅗×C = ⅗ × 5 = 3 D = 1⅖×C = 1⅖ × 5 = 7 and ABCDEF = 535750