Puzzle for April 22, 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 DE are 2-digit numbers (not B×C or D×E).
Scratchpad
Help Area
Hint #1
eq.6 may be written as: 10×B + C = 10×D + E + F In the above equation, replace E with C – F (from eq.2): 10×B + C = 10×D + C – F + F which becomes 10×B + C = 10×D + C Subtract C from both sides: 10×B + C – C = 10×D + C – C which becomes 10×B = 10×D Divide both sides by 10: 10×B ÷ 10 = 10×D ÷ 10 which makes B = D
Hint #2
Add A to both sides of eq.3: A – C + A = B – A + A which becomes eq.3a) 2×A – C = B Substitute 2×A – C for B (from eq.3a) in eq.5: C + F = A + 2×A – C – C which becomes C + F = 3×A – 2×C Add 2×C to both sides of the above equation: C + F + 2×C = 3×A – 2×C + 2×C which becomes eq.5a) 3×C + F = 3×A
Hint #3
In eq.4, substitute B for D: B + F = A + C – F Add F to both sides of the above equation: B + F + F = A + C – F + F which becomes eq.4a) B + 2×F = A + C
Hint #4
Substitute 2×A – C for B (from eq.3a) in eq.4a: 2×A – C + 2×F = A + C In the above equation, subtract A and 2×F from both sides, and add C to both sides: 2×A – C + 2×F – A – 2×F + C = A + C – A – 2×F + C which simplifies to eq.4b) A = 2×C – 2×F
Hint #5
Substitute (2×C – 2×F) for A (from eq.4b) into eq.5a: 3×C + F = 3×(2×C – 2×F) which becomes 3×C + F = 6×C – 6×F In the above equation, subtract 3×C from both sides, and add 6×F to both sides: 3×C + F – 3×C + 6×F = 6×C – 6×F – 3×C + 6×F which simplifies to 7×F = 3×C Divide both sides by 3: 7×F ÷ 3 = 3×C ÷ 3 which makes 2⅓×F = C
Hint #6
Substitute (2⅓×F) for C in eq.4b: A = 2×(2⅓×F) – 2×F which becomes A = 4⅔×F – 2×F which makes A = 2⅔×F
Hint #7
Substitute 2⅔×F for A, and 2⅓×F for C in eq.3a: 2×(2⅔×F) – 2⅓×F = B which becomes 5⅓×F – 2⅓×F = B which makes 3×F = B and also makes 3×F = B = D
Hint #8
Substitute (2⅓×F) for C in eq.2: E = 2⅓×F – F which makes E = 1⅓×F
Solution
Substitute 2⅔×F for A, 3×F for B and D, 2⅓×F for C, and 1⅓×F for E in eq.1: 2⅔×F + 3×F + 2⅓×F + 3×F + 1⅓×F + F = 40 which simplifies to 13⅓×F = 40 Divide both sides of the above equation by 13⅓: 13⅓×F ÷ 13⅓ = 40 ÷ 13⅓ which means F = 3 making A = 2⅔×F = 2⅔ × 3 = 8 B = D = 3×F = 3 × 3 = 9 C = 2⅓×F = 2⅓ × 3 = 7 E = 1⅓×F = 1⅓ × 3 = 4 and ABCDEF = 897943