Puzzle for May 19, 2019 ( )
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.
* AB, BC, CD, and EF are 2-digit numbers (not A×B, B×C, C×D, or E×F).
Scratchpad
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Hint #1
Add F to both sides of eq.4: B + C + D + F = A - B + E + F In the above equation, replace C + D + F with A + B (from eq.3): B + A + B = A - B + E + F Subtract A from both sides, and add B to both sides: B + A + B - A + B = A - B + E + F - A + B which becomes eq.3a) 3×B = E + F
Hint #2
Subtract the left and right sides of eq.4 from the left and right sides of eq.2, respectively: B + C + E - (B + C + D) = A + D + F - (A - B + E) which is the same as B + C + E - B - C - D = A + D + F - A + B - E which becomes E - D = D + F + B - E Add D and E to each side of the equation above: E - D + D + E = D + F + B - E + D + E which becomes 2×E = 2×D + B + F Multiply both sides by 3: 3×(2×E) = 3×(2×D + B + F) which is equivalent to eq.4a) 6×E = 6×D + 3×B + 3×F
Hint #3
In eq.4a, replace 3×B with E + F (from eq.3a): 6×E = 6×D + E + F + 3×F which becomes 6×E = 6×D + E + 4×F Subtract E and 4×F from both sides: 6×E - E - 4×F = 6×D + E + 4×F - E - 4×F which becomes eq.4b) 5×E - 4×F = 6×D
Hint #4
In eq.6, replace EF with AB + C + E (from eq.5): AB + CD = BC - D + AB + C + E Subtract AB from each side of the above equation: AB + CD - AB = BC - D + AB + C + E - AB which becomes CD = BC - D + C + E which may be written as 10×C + D = 10×B + C - D + C + E which becomes eq.6a) 10×C + D = 10×B + 2×C - D + E
Hint #5
In eq.6a, add D to both sides, and subtract 2×C from both sides: 10×C + D + D - 2×C = 10×B + C - D + C + E + D - 2×C which becomes 8×C + 2×D = 10×B + E Multiply both sides of the above equation by 3: 3×(8×C + 2×D) = 3×(10×B + E) which is the same as eq.6b) 24×C + 6×D = 10×(3×B) + 3×E
Hint #6
In eq.6b, substitute 5×E - 4×F for 6×D (from eq.4b), and E + F for 3×B (from eq.3a): 24×C + 5×E - 4×F = 10×(E + F) + 3×E which may be written as 24×C + 5×E - 4×F = 10×E + 10×F + 3×E Subtract 5×E from each side, and add 4×F to each side: 24×C + 5×E - 4×F - 5×E + 4×F = 10×E + 10×F + 3×E - 5×E + 4×F which simplifies to 24×C = 8×E + 14×F Divide both sides by 2: 24×C ÷ 2 = (8×E + 14×F) ÷ 2 which becomes eq.6c) 12×C = 4×E + 7×F
Hint #7
Multiply both sides of eq.3 by 12: 12×(C + D + F) = 12×(A + B) which is equivalent to 12×C + 12×D + 12×F = 12×A + 12×B which may be written as eq.3b) 12×C + 2×(6×D) + 12×F = 12×A + 4×(3×B)
Hint #8
Substitute 4×E + 7×F for 12×C (from eq.6c), 5×E - 4×F for 6×D (from eq.4b), and E + F for 3×B (from eq.3a) in eq.3b: 4×E + 7×F + 2×(5×E - 4×F) + 12×F = 12×A + 4×(E + F) which is equivalent to 4×E + 7×F + 10×E - 8×F + 12×F = 12×A + 4×E + 4×F which becomes 14×E + 11×F = 12×A + 4×E + 4×F Subtract both 4×E and 4×F from each side: 14×E + 11×F - 4×E - 4×F = 12×A + 4×E + 4×F - 4×E - 4×F which becomes eq.3c) 10×E + 7×F = 12×A
Hint #9
eq.5 may be written as: 10×E + F = 10×A + B + C + E Multiply both sides of the above equation by 12: 12×(10×E + F) = 12×(10×A + B + C + E) which becomes 120×E + 12×F = 120×A + 12×B + 12×C + 12×E which may be written as eq.5a) 120×E + 12×F = 10×(12×A) + 4×(3×B) + 12×C + 12×E
Hint #10
Substitute 10×E + 7×F for 12×A (from eq.3c), E + F for 3×B (from eq.3a), and 4×E + 7×F for 12×C (from eq.6c) in eq.5a: 120×E + 12×F = 10×(10×E + 7×F) + 4×(E + F) + 4×E + 7×F + 12×E which is the same as 120×E + 12×F = 100×E + 70×F + 4×E + 4×F + 16×E + 7×F which becomes 120×E + 12×F = 120×E + 81×F Subtract both 120×E and 12×F from both sides: 120×E + 12×F - 120×E - 12×F = 120×E + 81×F - 120×E - 12×F which simplifies to 0 = 69×F which means 0 = F
Hint #11
Substitute 0 for F in eq.3c: 10×E + 7×0 = 12×A which means 10×E = 12×A Divide both sides by 12: 10×E ÷ 12 = 12×A ÷ 12 which makes ⅚×E = A
Hint #12
Substitute 0 for F in eq.3a: 3×B = E + 0 which means 3×B = E Divide both sides by 3: 3×B ÷ 3 = E ÷ 3 which makes B = ⅓×E
Hint #13
Substitute 0 for F in eq.6c: 12×C = 4×E + 7×0 which means 12×C = 4×E Divide both sides by 12: 12×C ÷ 12 = 4×E ÷ 12 which makes C = ⅓×E
Hint #14
Substitute 0 for F in eq.4b: 5×E - 4×0 = 6×D which means 5×E = 6×D Divide both sides by 6: 5×E ÷ 6 = 6×D ÷ 6 which makes ⅚×E = D
Solution
Substitute ⅚×E for A and D, ⅓×E for B and C, and 0 for F in eq.1: ⅚×E + ⅓×E + ⅓×E + ⅚×E + E + 0 = 20 which simplifies to 3⅓×E = 20 Divide both sides by 3⅓: 3⅓×E ÷ 3⅓ = 20 ÷ 3⅓ which means E = 6 making A = D = ⅚×E = ⅚ × 6 = 5 B = C = ⅓×E = ⅓ × 6 = 2 and ABCDEF = 522560