Puzzle for January 2, 2022 ( )
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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 is a 2-digit number (not B×C).
** "F ^ E" means "F raised to the power of E".
*** "C mod F" equals the remainder of (C ÷ F).
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Hint #1
Add E to both sides of eq.2: C – E + E = E + F + E which becomes eq.2a) C = 2×E + F Since C, E, and F are all non-negative integers, then eq.2a implies that: C ≥ F
Hint #2
eq.6 may be expressed as: eq.6a) E = remainder of (C ÷ F) which may be written as eq.6b) (n × F) + E = C where n is an integer ≥ 0 In a division problem, the remainder will be less than the divisor. Therefore, from eq.6a, this makes: E < F
Hint #3
In eq.2a, replace C with (n × F) + E (from eq.6b): (n × F) + E = 2×E + F Subtract E and F from each side of the above equation: (n × F) + E – E – F = 2×E + F – E – F which becomes (n × F) – F = E which may be written as eq.2b) (n – 1) × F = E
Hint #4
To make eq.2b true, check several possible values for n: If n = 0, then E = (0 – 1) × F = (–1) × F = –F If n = 1, then E = (1 – 1) × F = 0 × F = 0 If n = 2, then E = (2 – 1) × F = 1 × F = F If n = 3, then E = (3 – 1) × F = 2 × F = 2×F If n > 3, then E > 2×F Since E must be a non-negative integer, and also since E < F, then E = 0 which makes n = 1
Hint #5
In eq.2a, replace E with 0: C = 2×0 + F which makes C = F
Hint #6
In eq.1, replace E with 0: A – 0 = B – A + 0 which becomes A = B – A Add A to both sides of the equation above: A + A = B – A + A which makes eq.1a) 2×A = B
Hint #7
In eq.5, substitute 2×A for B, and 0 for E: 2×A ÷ A = (D ÷ C) + (F ^ 0) which becomes 2 = (D ÷ C) + 1 Subtract 1 from each side of the equation above: 2 – 1 = (D ÷ C) + 1 – 1 which becomes 1 = D ÷ C Multiply both sides by C: 1 × C = D ÷ C × C which makes C = D
Hint #8
Substitute C for F in eq.3: D – B = C ÷ C which becomes D – B = 1 Add B to both sides of the above equation: D – B + B = 1 + B which becomes eq.3a) D = 1 + B
Hint #9
eq.4 may be written as: 10×B + C = B + (C×D) Substitute (1 + B) for D (from eq.3a) into the above equation: 10×B + C = B + (C×(1 + B)) which becomes 10×B + C = B + C + C×B Subtract C and B from both sides: 10×B + C – B – C = B + C + C×B – B – C which makes eq.4a) 9×B = C×B
Hint #10
Since A ≠ 0 (from eq.5), then from eq.1a: 2×A = B which implies B ≠ 0 Since B ≠ 0, divide both sides of eq.4a by B: 9×B ÷ B = C×B ÷ B which makes 9 = C and also makes F = C = D = 9
Hint #11
Substitute 9 for D in eq.3a: 9 = 1 + B Subtract 1 from each side of the above equation: 9 – 1 = 1 + B – 1 which makes 8 = B
Solution
Substitute 8 for B in eq.1a: 2×A = 8 Divide both sides of the above equation by 2: 2×A ÷ 2 = 8 ÷ 2 which makes A = 4 and makes ABCDEF = 489909