Question

how many numbers are there between 2000 and 3000 that are divisible by 9

Answer 1

Numbers which are divisible by 3

2001,2004,2007……………..2997————->A.S(Arithmetic series)

first term = a = 2001

Common difference = d = T2–T1 = 2001–2004 = 3

Last term = Tn = 2997

Number of terms = n = ?

Tn = a+(n-1)d

Tn-a+d/d = n

2997–2001+3/3 = n

333 = n

So , there are 333 terms which are divisible by 3

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2004,2008,2012,……………….2996———-> A.S

a = 2004

d = T2-T1 = 2008–2004 = 4

Last term = Tn = 2996

Number of terms = n = ?

Tn = a+(n-1)d

Tn-a+d/d = n

2996–2004+4/4 = n

249 = n

So there are 249 numbers which are Divisible by 4