Puzzle for April 2, 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.
Scratchpad
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Hint #1
In eq.3, subtract C from each side: A + C + D – C = B + E – C which becomes eq.3a) A + D = B + E – C In eq.2, add D to both sides, and subtract F from both sides: B + C – D + D – F = A + F + D – F which becomes eq.2a) B + C – F = A + D
Hint #2
In eq.3a, replace A + D with B + C – F (from eq.2a): B + C – F = B + E – C In the equation above, subtract B from both sides, and add C to both sides: B + C – F – B + C = B + E – C – B + C which becomes eq.3b) 2×C – F = E
Hint #3
eq.6 may be written as: F = (A + B + C + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + B + C + E) ÷ 4 which becomes eq.6a) 4×F = A + B + C + E Add C, F, and A to both sides of eq.5: B – C – F + C + F + A = C + F – A + C + F + A which simplifies to B + A = 2×C + 2×F which is the same as eq.5a) A + B = 2×C + 2×F
Hint #4
In eq.6a, replace A + B with 2×C + 2×F (from eq.5a): 4×F = 2×C + 2×F + C + E which becomes 4×F = 3×C + 2×F + E Subtract 2×F from each side of the equation above: 4×F – 2×F = 3×C + 2×F + E – 2×F which becomes eq.6b) 2×F = 3×C + E
Hint #5
In eq.6b, substitute 2×C – F for E (from eq.3b): 2×F = 3×C + 2×C – F which becomes 2×F = 5×C – F Add F to both sides of the equation above: 2×F + F = 5×C – F + F which makes 3×F = 5×C Divide both sides by 3: 3×F ÷ 3 = 5×C ÷ 3 which makes F = 1⅔×C
Hint #6
Substitute 1⅔×C for F in eq.3b: 2×C – 1⅔×C = E which makes ⅓×C = E
Hint #7
Substitute 1⅔×C for F, and ⅓×C for E in eq.4: C + 1⅔×C = A + D + ⅓×C which becomes 2⅔×C = A + D + ⅓×C Subtract ⅓×C from each side of the above equation: 2⅔×C – ⅓×C = A + D + ⅓×C – ⅓×C which becomes eq.4a) 2⅓×C = A + D
Hint #8
eq.3 may be written as: A + D + C = B + E Substitute 2⅓×C for A + D (from eq.4a), and ⅓×C for E into the equation above: 2⅓×C + C = B + ⅓×C which becomes 3⅓×C = B + ⅓×C Subtract ⅓×C from each side: 3⅓×C – ⅓×C = B + ⅓×C – ⅓×C which makes 3×C = B
Hint #9
Substitute 3×C for B, and (1⅔×C) for F in eq.5a: A + 3×C = 2×C + 2×(1⅔×C) which becomes A + 3×C = 2×C + 3⅓×C which becomes A + 3×C = 5⅓×C Subtract 3×C from each side of the above equation: A + 3×C – 3×C = 5⅓×C – 3×C which makes A = 2⅓×C
Hint #10
Substitute 2⅓×C for A in eq.4a: 2⅓×C = 2⅓×C + D Subtract 2⅓×C from each side of the above equation: 2⅓×C – 2⅓×C = 2⅓×C + D – 2⅓×C which means 0 = D
Solution
Substitute 2⅓×C for A, 3×C for B, 0 for D, ⅓×C for E, and 1⅔×C for F in eq.1: 2⅓×C + 3×C + C + 0 + ⅓×C + 1⅔×C = 25 which simplifies to 8⅓×C = 25 Divide both sides of the above equation by 8⅓: 8⅓×C ÷ 8⅓ = 25 ÷ 8⅓ which means C = 3 making A = 2⅓×C = 2⅓ × 3 = 7 B = 3×C = 3 × 3 = 9 E = ⅓×C = ⅓ × 3 = 1 F = 1⅔×C = 1⅔ × 3 = 5 and ABCDEF = 793015