Puzzle for January 16, 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.
* BC, CD, and AB are 2-digit numbers (not B×C, C×D, or A×B).
** B! is B-factorial.
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Hint #1
Add D to both sides of eq.1: C + D = B – D + D which becomes eq.1a) C + D = B eq.3 may be written as: 10×B + C = 10×C + D + E Subtract C from both sides of the above equation: 10×B + C – C = 10×C + D + E – C which becomes eq.3a) 10×B = 9×C + D + E
Hint #2
In eq.3a, substitute (C + D) for B (from eq.1a): 10×(C + D) = 9×C + D + E which becomes 10×C + 10×D = 9×C + D + E Subtract 9×C and D from each side of the above equation: 10×C + 10×D – 9×C – D = 9×C + D + E – 9×C – D which becomes eq.3b) C + 9×D = E
Hint #3
Since C ≠ 0 (from eq.2), then C ≥ 1 Since E is a one-digit non-negative integer, then 0 ≤ E ≤ 9 Combining these two inequalities with eq.3b yields: ie.3b) 1 ≤ C + 9×D ≤ 9
Hint #4
To make ie.3b true, check several possible values for D: If D = 0, then 1 ≤ C + 9×0 ≤ 9 making 1 ≤ C ≤ 9 If D = 1, then 1 ≤ C + 9×1 ≤ 9 making 1 ≤ C + 9 ≤ 9 which would make C = 0 If D = 2, then 1 ≤ C + 9×2 ≤ 9 making 1 ≤ C + 18 ≤ 9 which would make C ≤ –9 If D > 2, then C < –9 Since C ≥ 1, the above inequalities make D = 0
Hint #5
In eq.1, replace D with 0: C = B – 0 which makes C = B
Hint #6
In eq.3a, substitute B for C, and 0 for D: 10×B = 9×B + 0 + E which becomes 10×B = 9×B + E Subtract 9×B from both sides of the equation above: 10×B – 9×B = 9×B + E – 9×B which makes B = E
Hint #7
In eq.2, substitute B for C: F ÷ A = B ÷ B which becomes F ÷ A = 1 Multiply both sides of the above equation by A: F ÷ A × A = 1 × A which makes F = A
Hint #8
eq.4 may be written as: 10×A + B = 10×C + D + (E × F) Substitute B for C and E, 0 for D, and A for F in the above equation: 10×A + B = 10×B + 0 + (B × A) which becomes eq.4a) 10×A + B = 10×B + (B × A)
Hint #9
Subtract both B and (B × A) from both sides of eq.4a: 10×A + B – B – (B × A) = 10×B + (B × A) – B – (B × A) which becomes 10×A – (B × A) = 9×B which may be written as A×(10 – B) = 9×B Divide both sides of the above equation by (10 – B): A×(10 – B) ÷ (10 – B) = 9×B ÷ (10 – B) which becomes eq.4b) A = 9×B ÷ (10 – B)
Hint #10
Substitute B for E and C in eq.5: B! + B = B × B × B which may be written as B! + B = B³ Subtract B from each side of the above equation: B! + B – B = B³ – B which becomes B! = B³ – B which may be written as eq.5a) B! = B × (B² – 1)
Hint #11
eq.5a may be re-written as: 1 × 2 × 3 × ... × (B – 2) × (B – 1) × B = B × (B² – 1) which is equivalent to (B – 1)! × B = B × (B² – 1) Since B ≠ 0 (B = C, and C ≠ 0), divide both sides of the above equation by B: (B – 1)! × B ÷ B = B × (B² – 1) ÷ B which becomes (B – 1)! = B² – 1 which is equivalent to (B – 2)! × (B – 1) = (B + 1) × (B – 1) Divide both sides by (B – 1) (assumes B ≠ 1): (B – 2)! × (B – 1) ÷ (B – 1) = (B + 1) × (B – 1) ÷ (B – 1) which becomes eq.5b) (B – 2)! = B + 1
Hint #12
Confirm: B ≠ 1 ... Substituting 1 for B, C, and E in eq.5 would yield: 1! + 1 = 1 × 1 × 1 which would make 1 + 1 = 1 which would mean 2 = 1 Since 2 ≠ 1, then: B ≠ 1
Hint #13
Re-write eq.5b by subtracting (B – 2)! from each side: (B – 2)! – (B – 2)! = B + 1 – (B – 2)! which becomes eq.5c) 0 = B + 1 – (B – 2)! To make eq.5c true, check several possible values for B (B ≥ 2): If B = 2, then 0 = 2 + 1 – (2 – 2)! = 3 – (0)! = 3 – 1 = 2 If B = 3, then 0 = 3 + 1 – (3 – 2)! = 4 – (1)! = 4 – 1 = 3 If B = 4, then 0 = 4 + 1 – (4 – 2)! = 5 – (2)! = 5 – 2 = 3 If B = 5, then 0 = 5 + 1 – (5 – 2)! = 6 – (3)! = 6 – 6 = 0 If B = 6, then 0 = 6 + 1 – (6 – 2)! = 7 – (4)! = 7 – 24 = –17 If B > 6, then B + 1 – (B – 2)! < –17 Therefore, the only value for B that makes eq.5c true is: B = 5 which means C = B = E = 5
Solution
Substitute 5 for B in eq.4b: A = 9×5 ÷ (10 – 5) which becomes A = 45 ÷ 5 which makes A = 9 and also makes F = A = 9 and makes ABCDEF = 955059