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[2015 solutions]
... 3. We know that the sequence 1/n converges to zero. Since a < b, we can get N so that 1/N < b − a. Consider the numbers {k/N } where k runs over all integers, positive as well as negative. k/N decreases to −∞ as k decreases to −∞ and k/N increases to +∞ as k increases to ∞. Let K be the largest inte ...
... 3. We know that the sequence 1/n converges to zero. Since a < b, we can get N so that 1/N < b − a. Consider the numbers {k/N } where k runs over all integers, positive as well as negative. k/N decreases to −∞ as k decreases to −∞ and k/N increases to +∞ as k increases to ∞. Let K be the largest inte ...
The number of rational numbers determined by large sets of integers
... in [1, X] and [1, Y ] respectively, satisfying |A| αX and |B| βY , where X, Y real numbers at least 1, α, β are real numbers in (0, 1]. A standard application of the Möbius inversion formula then shows that |A/B| αβXY . Our purpose is to investigate what might be deduced when in place of inte ...
... in [1, X] and [1, Y ] respectively, satisfying |A| αX and |B| βY , where X, Y real numbers at least 1, α, β are real numbers in (0, 1]. A standard application of the Möbius inversion formula then shows that |A/B| αβXY . Our purpose is to investigate what might be deduced when in place of inte ...
Quiz 2 Solutions
... 177 -> Divisible by three b.c. the sum of digits = 9 = 3*59 -> 59 is prime so we're done ...
... 177 -> Divisible by three b.c. the sum of digits = 9 = 3*59 -> 59 is prime so we're done ...
Talent 02III
... 5. Alice and Bob play a game by taking turns removing some stones from a pile. The rules require that the number of stones removed at each turn must be a divisor of the number of stones in the pile at the start of that turn, and no player is ever allowed to take all of the stones. The winner of the ...
... 5. Alice and Bob play a game by taking turns removing some stones from a pile. The rules require that the number of stones removed at each turn must be a divisor of the number of stones in the pile at the start of that turn, and no player is ever allowed to take all of the stones. The winner of the ...
Full text
... of all the points for a given n yields hn. Both Dickson [3] and LeVeque [4] provide reviews concerning heptagonal and related figurate numbers. Our analysis starts with the observation from a table of hn values [5] that h17 = he + h165 h58 = hl:L + h5?9 and \ l h = hl6 + fc123. Note that each of the ...
... of all the points for a given n yields hn. Both Dickson [3] and LeVeque [4] provide reviews concerning heptagonal and related figurate numbers. Our analysis starts with the observation from a table of hn values [5] that h17 = he + h165 h58 = hl:L + h5?9 and \ l h = hl6 + fc123. Note that each of the ...
