Puzzle for September 1, 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 DE are 2-digit numbers (not A×B, B×C, C×D, or D×E).
** BCD and DEF are 3-digit numbers (not B×C×D or D×E×F).
Scratchpad
Help Area
Hint #1
eq.4 may be written as: 10×D + E = 10×B + C + B which becomes eq.4a) 10×D + E = 11×B + C eq.6 may be written as: B + C + D + 100×D + 10×E + F = 100×B + 10×C + D + 10×D + E which becomes B + C + 101×D + 10×E + F = 100×B + 10×C + 11×D + E Subtract B, C, 11×D, and E from each side of the equation above: B + C + 101×D + 10×E + F – B – C – 11×D – E = 100×B + 10×C + 11×D + E – B – C – 11×D – E which simplifies to eq.6a) 90×D + 9×E + F = 99×B + 9×C
Hint #2
Multiply both sides of eq.4a by 9: 9×(10×D + E) = 9×(11×B + C) which becomes 90×D + 9×E = 99×B + 9×C In the above equation, replace 99×B + 9×C with 90×D + 9×E + F (from eq.6a): 90×D + 9×E = 90×D + 9×E + F Subtract 90×D and 9×E from both sides of the equation above: 90×D + 9×E – 90×D – 9×E = 90×D + 9×E + F – 90×D – 9×E which simplifies to 0 = F
Hint #3
In eq.2, add B to both sides, and subtract D from each side: B + C + B – D = A – B + D + B – D which becomes 2×B + C – D = A In eq.3, substitute 2×B + C – D for A, and 0 for F: 2×B + C – D = B + D + E – 0 In the above equation, add D to both sides, and subtract B from both sides: 2×B + C – D + D – B = B + D + E – 0 + D – B which becomes eq.3a) B + C = 2×D + E
Hint #4
eq.4a may be written as: 8×D + 2×D + E = 11×B + C In the above equation, substitute B + C for 2×D + E (from eq.3a): 8×D + B + C = 11×B + C Subtract B and C from both sides: 8×D + B + C – B – C = 11×B + C – B – C which becomes 8×D = 10×B Divide both sides by 8: 8×D ÷ 8 = 10×B ÷ 8 which becomes D = 1¼×B
Hint #5
Substitute (1¼×B) for D in eq.3a: B + C = 2×(1¼×B) + E which becomes B + C = 2½×B + E Subtract B from each side of the above equation: B + C – B = 2½×B + E – B which becomes eq.4b) C = 1½×B + E
Hint #6
Substitute 1¼×B for D, and 0 for F in eq.3: A = B + 1¼×B + E – 0 which becomes eq.3b) A = 2¼×B + E
Hint #7
eq.5 may be written as: 10×C + D + A + B + C + D + F = 10×A + B – D which becomes 11×C + 2×D + A + B + F = 10×A + B – D In the above equation, add D to both sides, and subtract A and B from both sides: 11×C + 2×D + A + B + F + D – A – B = 10×A + B – D + D – A – B which simplifies to eq.5a) 11×C + 3×D + F = 9×A
Hint #8
Substitute (1½×B + E) for C (from eq.4b), (1¼×B) for D, 0 for F, and (2¼×B + E) for A (from eq.3b) in eq.5a: 11×(1½×B + E) + 3×(1¼×B) + 0 = 9×(2¼×B + E) which is equivalent to 16½×B + 11×E + 3¾×B = 20¼×B + 9×E which becomes 20¼×B + 11×E = 20¼×B + 9×E Subtract 20¼×B and 9×E from each side of the equation above: 20¼×B + 11×E – 20¼×B – 9×E = 20¼×B + 9×E – 20¼×B – 9×E which simplifies to 0 + 2×E = 0 which means E = 0
Hint #9
Substitute 0 for E in eq.3b: A = 2¼×B + 0 which makes A = 2¼×B
Hint #10
Substitute 0 for E in eq.4b: C = 1½×B + 0 which makes C = 1½×B
Solution
Substitute 2¼×B for A, 1½×B for C, 1¼×B for D, and 0 for E and F in eq.1: 2¼×B + B + 1½×B + 1¼×B + 0 + 0 = 24 which simplifies to 6×B = 24 Divide both sides of the equation above by 6: 6×B ÷ 6 = 24 ÷ 6 which means B = 4 making A = 2¼×B = 2¼ × 4 = 9 C = 1½×B = 1½ × 4 = 6 D = 1¼×B = 1¼ × 4 = 5 and ABCDEF = 946500