Puzzle for May 28, 2023 ( )
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.
* CD is a 2-digit number (not C×D).
Scratchpad
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Hint #1
eq.4 may be written as: 10×C + D - F = A + B + C + F In the above equation, add F to both sides, and subtract C from both sides: 10×C + D - F + F - C = A + B + C + F + F - C which becomes 9×C + D = A + B + 2×F which may be written as eq.4a) 9×C + D = A + F + B + F
Hint #2
In eq.4a, replace B + F with A + D (from eq.2): 9×C + D = A + F + A + D which becomes 9×C + D = 2×A + F + D Subtract D and 2×A from each side of the equation above: 9×C + D - D - 2×A = 2×A + F + D - D - 2×A which becomes eq.4b) 9×C - 2×A = F
Hint #3
eq.5 may be written as: E - C = (A + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (E - C) = 3 × (A + D + F) ÷ 3 which becomes eq.5a) 3×E - 3×C = A + D + F
Hint #4
In eq.5a, replace A + D with B + F (from eq.2): 3×E - 3×C = B + F + F which becomes eq.5b) 3×E - 3×C = B + 2×F
Hint #5
Add F and C to both sides of eq.3: C + E - F + F + C = B - C + D + F + C which becomes eq.3a) 2×C + E = B + D + F
Hint #6
Subtract the left and right sides of eq.3a from the left and right sides of eq.5a, respectively: 3×E - 3×C - (2×C + E) = A + D + F - (B + D + F) which becomes 3×E - 3×C - 2×C - E = A + D + F - B - D - F which becomes eq.5c) 2×E - 5×C = A - B
Hint #7
In eq.5b, substitute (9×C + 2×A) for F (from eq.4b): 3×E - 3×C = B + 2×(9×C - 2×A) which becomes 3×E - 3×C = B + 18×C - 4×A Add 3×C and 4×A to both sides of the above equation: 3×E - 3×C + 3×C + 4×A = B + 18×C - 4×A + 3×C + 4×A which becomes eq.5d) 3×E + 4×A = B + 21×C
Hint #8
Add the left and right sides of eq.5c to the left and right sides of eq.5d, respectively: 3×E + 4×A + 2×E - 5×C = B + 21×C + A - B which becomes 5×E + 4×A - 5×C = 21×C + A In the above equation, add 5×C to both sides, and subtract A from both sides: 5×E + 4×A - 5×C + 5×C - A = 21×C + A + 5×C - A which becomes eq.5e) 5×E + 3×A = 26×C
Hint #9
Multiply both sides of eq.6 by C: C × (C + (E ÷ C)) = C × (B ÷ C) which becomes eq.6a) C² + E = B
Hint #10
Substitute (C² + E) for B (from eq.6a) in eq.5c: 2×E - 5×C = A - (C² + E) which becomes 2×E - 5×C = A - C² - E Add C² and E to both sides of the above equation: 2×E - 5×C + C² + E = A - C² - E + C² + E which becomes eq.3b) 3×E - 5×C + C² = A
Hint #11
Substitute (3×E - 5×C + C²) for A (from eq.3b) in eq.5e: 5×E + 3×(3×E - 5×C + C²) = 26×C which becomes 5×E + 9×E - 15×C + 3×C² = 26×C which becomes 14×E - 15×C + 3×C² = 26×C In the equation above, add 15×C to both sides, and subtract 3×C² from both sides: 14×E - 15×C + 3×C² + 15×C - 3×C² = 26×C + 15×C - 3×C² which becomes eq.5f) 14×E = 41×C - 3×C²
Hint #12
Multiply both sides of eq.6a by 14: 14×(C² + E) = 14×B which becomes 14×C² + 14×E = 14×B In the equation above, substitute 41×C - 3×C² for 14×E (from eq.5f): 14×C² + 41×C - 3×C² = 14×B which becomes eq.6b) 11×C² + 41×C = 14×B
Hint #13
Multiply both sides of eq.3b by 14: 14×(3×E - 5×C + C²) = 14×A which becomes 14×(3×E) - 70×C + 14×C² = 14×A which may be written as 3×(14×E) - 70×C + 14×C² = 14×A Substitute 41×C - 3×C² for 14×E (from eq.5f) in the above equation: 3×(41×C - 3×C²) - 70×C + 14×C² = 14×A which becomes 123×C - 9×C² - 70×C + 14×C² = 14×A which becomes eq.3c) 53×C + 5×C² = 14×A
Hint #14
Multiply both sides of eq.4b by 14: 14×(9×C - 2×A) = 14×F which becomes 126×C - 14×(2×A) = 14×F which may be written as 126×C - 2×(14×A) = 14×F Substitute 53×C + 5×C² for 14×A (from eq.3c) in the above equation: 126×C - 2×(53×C + 5×C²) = 14×F which becomes 126×C - 106×C - 10×C²) = 14×F which becomes eq.4c) 20×C - 10×C² = 14×F
Hint #15
Multiply both sides of eq.2 by 14: 14×(A + D) = 14×(B + F) which becomes 14×A + 14×D = 14×B + 14×F Substitute 53×C + 5×C² for 14×A (from eq.3c), 11×C² + 41×C for 14×B (from eq.6b), and 20×C - 10×C² for 14×F (from eq.4c) in the above equation: 53×C + 5×C² + 14×D = 11×C² + 41×C + 20×C - 10×C² which becomes 53×C + 5×C² + 14×D = C² + 61×C Subtract 53×C and 5×C² from each side: 53×C + 5×C² + 14×D - 53×C - 5×C² = C² + 61×C - 53×C - 5×C² which becomes eq.2a) 14×D = 8×C - 4×C²
Hint #16
Multiply both sides of eq.1 by 14: 14×(A + B + C + D + E + F) = 14×25 which becomes 14×A + 14×B + 14×C + 14×D + 14×E + 14×F = 350 In the above equation, substitute 53×C + 5×C² for 14×A (from eq.3c), 11×C² + 41×C for 14×B (from eq.6b), 8×C - 4×C² for 14×D (from eq.2a), 41×C - 3×C² for 14×E (from eq.5f), and 20×C - 10×C² for 14×F (from eq.4c): 53×C + 5×C² + 11×C² + 41×C + 14×C + 8×C - 4×C² + 41×C - 3×C² + 20×C - 10×C² = 350 which simplifies to 177×C - C² = 350 Subtract 177×C from both sides, and add C² to both sides: 177×C - C² - 177×C + C² = 350 - 177×C + C² which becomes 0 = 350 - 177×C + C² which may be written as eq.1a) 0 = C² - 177×C + 350
Hint #17
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.1a yields: C = { (-1)×(-177) ± sq.rt.[(-177)² - (4 × 1 × 350)] } ÷ (2 × 1) which becomes C = {177 ± sq.rt.(31329 - 1400)} ÷ 2 which becomes C = {177 ± sq.rt.(29929)} ÷ 2 which makes eq.1b) C = (177 ± 173) ÷ 2
Hint #18
In eq.1b, either C = (177 + 173) ÷ 2 = 350 ÷ 2 = 175 or C = (177 - 173) ÷ 2 = 4 ÷ 2 = 2 Since C must be a one-digit integer, then C ≠ 175 and therefore makes C = 2
Solution
Since C = 2, then: 14×A = 53×C + 5×C² = 53×2 + 5×2² = 106 + 5×4 = 106 + 20 = 126 which means A = 126 ÷ 14 = 9 (from eq.3c) 14×B = 11×C² + 41×C = 11×2² + 41×2 = 11×4 + 82 = 44 + 82 = 126 which means B = 126 ÷ 14 = 9 (from eq.6b) 14×D = 8×C - 4×C² = 8×2 - 4×2² = 16 - 4×4 = 16 - 16 = 0 which means D = 0 (from eq.2a) 14×E = 41×C - 3×C² = 41×2 - 3×2² = 82 - 3×4 = 82 - 12 = 70 which means E = 70 ÷ 14 = 5 (from eq.5f) 14×F = 20×C - 10×C² = 20×2 - 10×2² = 40 - 10×4 = 40 - 40 = 0 which means F = 0 (from eq.4c) and ABCDEF = 992050