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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} S=\{1,2,3,4,\cdots , 1000\} \hspace{.33em}\\~\\ \end{align}}\) How many arithmatic progressions can be formed from the elements of \(S\) that start with \(1\) and end with \(1000\) and have at least \(3\) elements.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0may u pls help me with my math?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, let's work on this problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(a_1=1\), \(a_n=1000\) let difference of terms be \(d\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u mean common difference

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we can calculate \(a_n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then, we have\[999=(n1)d\]so \(d\) must be a divisor of \(999\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, how many values \(d\) can take?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0considering\[999=3^3 \times 37\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mukushla can u help me pls

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mathmath333 can u pls help me with my question

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i didnt undestand about the value of d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well\[n1=\frac{999}{d}\]right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since \(n1\) is an integer so \(\frac{999}{d}\) must be an integer, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so d must be a divisor of 999

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, how many positive divisors does 999 have?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i forgot how to do it ,do we add the powers of its prime divisors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, that right,\[(3+1)(2+1)=8\]which are\[1, 3, 9, 27, 37, 111, 333, 999\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how many of them are acceptable considering conditions of problem?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i m still in doubt how u got this (3+1)(2+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sry, It's (3+1)(1+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0final answer will be 7, see if you can get this number as your final answer

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ans given is 7 , but how u got thaat

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well all of values for \(d\) are acceptable, except \(d=1\) which gives a value of \(n=2\), and our arithmatic progressions have at least 3 elements

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for example when you let \(d=333\) then \(n\) becomes:\[n1=\frac{999}{333}=3\]\[n=4\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u checked all 8 options

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no need to check all of them because it's obvious that all of values give a \(n\) that is acceptable

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0ohk thnx briliant/bright
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