Puzzle for March 20, 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.
* AB and DE are 2-digit numbers (not A×B or D×E).
Scratchpad
Help Area
Hint #1
Add E to both sides of eq.5: B + E + E = A + C – E + E which becomes eq.5a) B + 2×E = A + C Add A to both sides of eq.3: B + D + A = C + E + A which is the same as eq.3a) A + B + D = A + C + E
Hint #2
In eq.3a, replace A + C with B + 2×E (from eq.5a): A + B + D = B + 2×E + E which becomes A + B + D = B + 3×E Subtract B and D from each side of the above equation: A + B + D – B – D = B + 3×E – B – D which becomes eq.3b) A = 3×E – D
Hint #3
eq.6 may be re-written as: A – B + D = 10×A + B – (10×D + E) which becomes A – B + D = 10×A + B – 10×D – E In the above equation, subtract A from both sides, and add B to both sides: A – B + D – A + B = 10×A + B – 10×D – E – A + B which becomes D = 9×A + 2×B – 10×D – E Add 10×D and E to both sides: D + 10×D + E = 9×A + 2×B – 10×D – E + 10×D + E which becomes eq.6a) 11×D + E = 9×A + 2×B
Hint #4
In eq.6a, substitute (3×E – D) for A (from eq.3b): 11×D + E = 9×(3×E – D) + 2×B which becomes 11×D + E = 27×E – 9×D + 2×B In the above equation, subtract 27×E from both sides, and add 9×D to both sides: 11×D + E – 27×E + 9×D = 27×E – 9×D + 2×B – 27×E + 9×D which becomes 20×D – 26×E = 2×B Divide both sides by 2: (20×D – 26×E) ÷ 2 = 2×B ÷ 2 which becomes eq.6b) 10×D – 13×E = B
Hint #5
In eq.5a, replace B with 10×D – 13×E (from eq.6b), and replace A with 3×E – D (from eq.3b): 10×D – 13×E + 2×E = 3×E – D + C which becomes 10×D – 11×E = 3×E – D + C In the above equation, subtract 3×E from both sides, and add D to both sides: 10×D – 11×E – 3×E + D = 3×E – D + C – 3×E + D which becomes eq.5b) 11×D – 14×E = C
Hint #6
In eq.2, substitute (3×E – D) for A (from eq.3b), and (11×D – 14×E) for C (from eq.5b): F – (3×E – D) = (3×E – D) – (11×D – 14×E) which becomes F – 3×E + D = 3×E – D – 11×D + 14×E which becomes F – 3×E + D = 17×E – 12×D In the above equation, add 3×E to both sides, and subtract D from both sides: F – 3×E + D + 3×E – D = 17×E – 12×D + 3×E – D which becomes eq.2a) F = 20×E – 13×D
Hint #7
Add D and B to both sides of eq.4: F – D + D + B = C – B + D + B which becomes F + B = C + D Subtract the left and right sides of eq.3 from the left and right sides of the above equation, respectively: F + B – (B + D) = C + D – (C + E) which becomes F + B – B – D = C + D – C – E which becomes F – D = D – E Add D to both sides of the equation above: F – D + D = D – E + D which becomes eq.4a) F = 2×D – E
Hint #8
Substitute 2×D – E for F (from eq.4a) into eq.2a: 2×D – E = 20×E – 13×D Add E and 13×D to both sides of the above equation: 2×D – E + E + 13×D = 20×E – 13×D + E + 13×D which becomes 15×D = 21×E Divide both sides by 3: 15×D ÷ 3 = 21×E ÷ 3 which makes eq.2b) 5×D = 7×E
Hint #9
Multiply both sides of eq.3b by 5: 5×A = 5×(3×E – D) which becomes 5×A = 15×E – 5×D Substitute 7×E for 5×D (from eq.2b) into the equation above: 5×A = 15×E – 7×E which becomes 5×A = 8×E Divide both sides by 8: 5×A ÷ 8 = 8×E ÷ 8 which makes ⅝×A = E
Hint #10
Substitute (⅝×A) for E in eq.2b: 5×D = 7×(⅝×A) which becomes 5×D = 4⅜×A Divide both sides by 5: 5×D ÷ 5 = 4⅜×A ÷ 5 which becomes D = ⅞×A
Hint #11
Substitute (⅞×A) for D, and ⅝×A for E in eq.4a: F = 2×(⅞×A) – ⅝×A which becomes F = 1¾×A – ⅝×A which makes F = 1⅛×A
Hint #12
Substitute (⅞×A) for D, and (⅝×A) for E in eq.6b: 10×(⅞×A) – 13×(⅝×A) = B which becomes 8¾×A – 8⅛×A = B which makes ⅝×A = B
Hint #13
Substitute (⅞×A) for D, and (⅝×A) for E in eq.5b: 11×(⅞×A) – 14×(⅝×A) = C which becomes 9⅝×A – 8¾×A = C which makes ⅞×A = C
Solution
Substitute ⅝×A for B and E, ⅞×A for C and D, and 1⅛×A for F in eq.1: A + ⅝×A + ⅞×A + ⅞×A + ⅝×A + 1⅛×A = 41 which simplifies to 5⅛×A = 41 Divide both sides of the above equation by 5⅛: 5⅛×A ÷ 5⅛ = 41 ÷ 5⅛ which means A = 8 making B = E = ⅝×A = ⅝ × 8 = 5 C = D = ⅞×A = ⅞ × 8 = 7 F = 1⅛×A = 1⅛ × 8 = 9 and ABCDEF = 857759