Puzzle for May 17, 2020 ( )
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.
* E! is E-factorial. BC and CD are 2-digit numbers (not B×C or C×D).
Scratchpad
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Hint #1
In eq.5, replace B with C + F (from eq.3): C + F – D + E = A + F In the above equation, subtract F from each side, and add D to each side: C + F – D + E – F + D = A + F – F + D which becomes eq.5a) C + E = A + D
Hint #2
In eq.5a, replace C + E with D + F (from eq.2): D + F = A + D Subtract D from each side of the equation above: D + F – D = A + D – D which becomes F = A
Hint #3
In eq.1, substitute C + F for B (from eq.3): A + C + F = C + D Subtract C from each side of the above equation: A + C + F – C = C + D – C which becomes eq.1a) A + F = D
Hint #4
Substitute A for F in eq.1a: A + A = D which makes 2×A = D
Hint #5
Substitute 2×A for D, and A for F in eq.5: B – 2×A + E = A + A which becomes B – 2×A + E = 2×A In the above equation, add 2×A to both sides, and subtract E from both sides: B – 2×A + E + 2×A – E = 2×A + 2×A – E which becomes eq.5b) B = 4×A – E
Hint #6
Substitute 2×A for D, and A for F in eq.2: C + E = 2×A + A which becomes C + E = 3×A Subtract E from each side of the above equation: C + E – E = 3×A – E which becomes eq.2a) C = 3×A – E
Hint #7
Substitute 2×A for D, and 3×A – E for C (from eq.2a) in eq.4: 2×A + E = B + 3×A – E In the above equation, subtract 3×A from both sides, and add E to both sides: 2×A + E – 3×A + E = B + 3×A – E – 3×A + E which becomes eq.4a) 2×E – A = B
Hint #8
Substitute 2×E – A for B (from eq.4a) in eq.5b: 2×E – A = 4×A – E Add A and E to both sides of the above equation: 2×E – A + A + E = 4×A – E + A + E which becomes 3×E = 5×A Divide both sides by 3: 3×E ÷ 3 = 5×A ÷ 3 which makes eq.5c) E = 1⅔×A
Hint #9
Substitute 1⅔×A for E in eq.2a: C = 3×A – 1⅔×A which makes C = 1⅓×A
Hint #10
Substitute (1⅔×A) for E in eq.4a: 2×(1⅔×A) – A = B which becomes 3⅓×A – A = B which makes 2⅓×A = B
Hint #11
eq.6 may be written as: E! = 10×B + C + 10×C + D which becomes E! = 10×B + 11×C + D Substitute (2⅓×A) for B, (1⅓×A) for C, and 2×A for D in the above equation: E! = 10×(2⅓×A) + 11×(1⅓×A) + 2×A which becomes E! = 23⅓×A + 14⅔×A + 2×A which makes eq.6a) E! = 40×A
Hint #12
Divide both sides of eq.5c by 1⅔: E ÷ 1⅔ = 1⅔×A ÷ 1⅔ which makes ⅗×E = A Substitute (⅗×E) for A in eq.6a: E! = 40×(⅗×E) which makes eq.6b) E! = 24×E (implies E ≠ 0)
Solution
Divide both sides of eq.6b by E: E! ÷ E = 24×E ÷ E which becomes (E – 1)! = 24 which means E – 1 = 4 Add 1 to each side of the above equation: E – 1 + 1 = 4 + 1 which means E = 5 making A = ⅗×E = ⅗ × 5 = 3 B = 2⅓×A = 2⅓ × 3 = 7 C = 1⅓×A = 1⅓ × 3 = 4 D = 2×A = 2 × 3 = 6 F = A = 3 and ABCDEF = 374653