Question

How many integers are there between 300 and 600 that are divisible by 9?

Answer 1

Answer:

There are 33 integers between 300 and 600.

Solution:

The first integer after 300 that is divisible by 9 is 306 = First number of an Arithmetic Progression with common difference 9.

So, first number (a) = 306

Common difference (d) = 9

And the last integer of the progression is 594 = nth term

We know,

$\begin{array} { c } { \mathrm { T } _ { \mathrm { n } } = \mathrm { a } + ( \mathrm { n } - 1 ) \mathrm { d } } \\\\ { 594 = 306 + ( \mathrm { n } - 1 ) 9 } \\\\ { 594 - 306 = 9 \mathrm { n } - 9 } \\\\ { 288 = 9 \mathrm { n } - 9 } \\\\ { 9 \mathrm { n } = 288 + 9 } \\\\ { 9 \mathrm { n } = 297 } \\\\ { \mathrm { n } = \frac { 297 } { 9 } = 33 } \end{array}$

Hence, there are 33 integers between 300 and 600.

Answer 2

Answer:

$There\:are \: 33 \: integers\: between \:300\\and \:600\:that\:are\\divisible \:by \:9$

Step-by-step explanation:

$Integers \: between \: 300\\and \:600\:that\:are\\divisible \:by \:9 \:are \:306,\\315,...,594\:is \:an\:A.P$

$First\:term (a)=306$

$common\: difference (d)=9$

$n^{th}\:term =594$

$\implies a+(n-1)d=594$

$\implies 306+(n-1)9=594$

Divide each term by 9, we get

$\implies 34+n-1=66$

$\implies 33+n=66$

$\implies n = 66-33$

$\implies n = 33$

Therefore,

$There\:are \: 33 \: integers\: between \:300\\and \:600\:that\:are\\divisible \:by \:9$

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