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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align} S=\{1,2,3,4,\cdots , 1000\} \hspace{.33em}\\~\\ \end{align}}\) 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.
may u pls help me with my math?
ok, let's work on this problem

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\(a_1=1\), \(a_n=1000\) let difference of terms be \(d\)
u mean common difference
yeah
Now we can calculate \(a_n\)
\(a_n=1+(n-1)d\)
very good
then, we have\[999=(n-1)d\]so \(d\) must be a divisor of \(999\)
yeas
ok, how many values \(d\) can take?
considering\[999=3^3 \times 37\]
@mukushla can u help me pls
@mathmath333 can u pls help me with my question
i didnt undestand about the value of d
well\[n-1=\frac{999}{d}\]right?
yes
since \(n-1\) is an integer so \(\frac{999}{d}\) must be an integer, right?
yes
so d must be a divisor of 999
yes
ok, how many positive divisors does 999 have?
i forgot how to do it ,do we add the powers of its prime divisors
yeah, that right,\[(3+1)(2+1)=8\]which are\[1, 3, 9, 27, 37, 111, 333, 999\]
how many of them are acceptable considering conditions of problem?
i m still in doubt how u got this (3+1)(2+1)
sry, It's (3+1)(1+1)
ok np
final answer will be 7, see if you can get this number as your final answer
ans given is 7 , but how u got thaat
well 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
for example when you let \(d=333\) then \(n\) becomes:\[n-1=\frac{999}{333}=3\]\[n=4\]
which is acceptable
u checked all 8 options
no need to check all of them because it's obvious that all of values give a \(n\) that is acceptable
ohk thnx briliant/bright
np, anytime math

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