Puzzle for July 3, 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.
* DE and CD are 2-digit numbers (not D×E or C×D).
Scratchpad
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Hint #1
Add A and E to both sides of eq.2: C – A + A + E = A – E + A + E which becomes eq.2a) C + E = 2×A Add D and A to both sides of eq.4: A + B – D + E + D + A = C + D – A + D + A which becomes eq.4a) 2×A + B + E = C + 2×D
Hint #2
In eq.4a, replace 2×A with C + E (from eq.2a): C + E + B + E = C + 2×D which becomes C + 2×E + B = C + 2×D Subtract C and 2×E from both sides of the above equation: C + 2×E + B – C – 2×E = C + 2×D – C – 2×E which becomes eq.4b) B = 2×D – 2×E
Hint #3
eq.6 may be written as: A = (B + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + D + E) ÷ 3 which becomes eq.6a) 3×A = B + D + E In eq.6a, replace B with 2×D – 2×E (from eq.4b): 3×A = 2×D – 2×E + D + E which becomes eq.6b) 3×A = 3×D – E
Hint #4
eq.5 may be written as: 10×D + E = A + 10×C + D Subtract D from both sides of the above equation: 10×D + E – D = A + 10×C + D – D which becomes eq.5a) 9×D + E = A + 10×C Subtract E from both sides of eq.2a: C + E – E = 2×A – E which becomes eq.2b) C = 2×A – E
Hint #5
In eq.5a, substitute (2×A – E) for C (from eq.2b): 9×D + E = A + 10×(2×A – E) which becomes 9×D + E = A + 20×A – 10×E which becomes 9×D + E = 21×A – 10×E Add 10×E to both sides of the equation above: 9×D + E + 10×E = 21×A – 10×E + 10×E which becomes 9×D + 11×E = 21×A which may be written as eq.5b) 9×D + 11×E = 7×(3×A)
Hint #6
Substitute (3×D – E) for 3×A (from eq.6b) in eq.5b: 9×D + 11×E = 7×(3×D – E) which becomes 9×D + 11×E = 21×D – 7×E which becomes 18×E = 12×D Divide both sides of the above equation by 12: 18×E ÷ 12 = 12×D ÷ 12 which makes 1½×E = D
Hint #7
Substitute (1½×E) for D in eq.6b: 3×A = 3×(1½×E) – E which becomes 3×A = 4½×E – E which makes 3×A = 3½×E Divide both sides of the above equation by 3: 3×A ÷ 3 = 3½×E ÷ 3 which makes A = 1⅙×E
Hint #8
Substitute 1⅙×E for A in eq.2b: C = 2×(1⅙×E) – E which becomes C = 2⅓×E – E which makes C = 1⅓×E
Hint #9
Substitute (1½×E) for D in eq.4b: B = 2×(1½×E) – 2×E which becomes B = 3×E – 2×E which makes B = E
Hint #10
Substitute 1½×E for D, 1⅙×E for A, and 1⅓×E for C in eq.3: 1½×E + F = 1⅙×E + 1⅓×E which becomes 1½×E + F = 2½×E Subtract 1½×E from each side of the equation above: 1½×E + F – 1½×E = 2½×E – 1½×E which makes F = E
Solution
Substitute 1⅙×E for A, E for B and F, 1⅓×E for C, and 1½×E for D in eq.1: 1⅙×E + E + 1⅓×E + 1½×E + E + E = 42 which simplifies to 7×E = 42 Divide both sides of the above equation by 7: 7×E ÷ 7 = 42 ÷ 7 which means E = 6 making A = 1⅙×E = 1⅙ × 6 = 7 B = F = E = 6 C = 1⅓×E = 1⅓ × 6 = 8 D = 1½×E = 1½ × 6 = 9 and ABCDEF = 768966