Question

question

Answer 1

\(\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.

Answer 2

may u pls help me with my math?

Answer 3

ok, let's work on this problem

Answer 4

\(a_1=1\), \(a_n=1000\) let difference of terms be \(d\)

Answer 5

u mean common difference

Answer 6

yeah

Answer 7

Now we can calculate \(a_n\)

Answer 8

\(a_n=1+(n-1)d\)

Answer 9

very good

Answer 10

then, we have\[999=(n-1)d\]so \(d\) must be a divisor of \(999\)

Answer 11

yeas

Answer 12

ok, how many values \(d\) can take?

Answer 13

considering\[999=3^3 \times 37\]

Answer 14

@mukushla can u help me pls

Answer 15

@mathmath333 can u pls help me with my question

Answer 16

i didnt undestand about the value of d

Answer 17

well\[n-1=\frac{999}{d}\]right?

Answer 18

yes

Answer 19

since \(n-1\) is an integer so \(\frac{999}{d}\) must be an integer, right?

Answer 20

yes

Answer 21

so d must be a divisor of 999

Answer 22

yes

Answer 23

ok, how many positive divisors does 999 have?

Answer 24

i forgot how to do it ,do we add the powers of its prime divisors

Answer 25

yeah, that right,\[(3+1)(2+1)=8\]which are\[1, 3, 9, 27, 37, 111, 333, 999\]

Answer 26

how many of them are acceptable considering conditions of problem?

Answer 27

i m still in doubt how u got this (3+1)(2+1)

Answer 28

sry, It's (3+1)(1+1)

Answer 29

ok np

Answer 30

final answer will be 7, see if you can get this number as your final answer

Answer 31

ans given is 7 , but how u got thaat

Answer 32

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

Answer 33

for example when you let \(d=333\) then \(n\) becomes:\[n-1=\frac{999}{333}=3\]\[n=4\]

Answer 34

which is acceptable

Answer 35

u checked all 8 options

Answer 36

no need to check all of them because it's obvious that all of values give a \(n\) that is acceptable

Answer 37

ohk thnx briliant/bright

Answer 38

np, anytime math