Question

How many two-digit numbers are divisible by 6 ?

Answer 1

HEYAA..

Here is ur answer....

The 1st two digit no. which are divisibe by 6 is 12 and the last two digit no. is 96.

LET a=12, d=6 and $a_{n}$=96

Now,

$a_{n}$=a+(n-1)d

96=12+(n-1)6

84=(n-1)6

n-1=14

n=14+1

n=15

Hence total no. of two digit no. divisible by 6 is 15

Answer 2

$\red {\boxed{\huge \frak{✓ \: Hola! \: Mate \:✓ }}} \:$
$\underline {\bf{QUESTION} }: \\ \\ \blue{\tiny \bf{How \: many \: two-digit \: numbers \: are \: divisible \: by \: 6 ?}} \\ \\ \underline{ \bf{SOLUTION}} : \\ \\ \small{\bf{there \: are \: some \:no. \: which \: are \: divisible \: of \: 6}} \\ \\ 12,18,24,.......96. \\ \\ \underline {\bf{As \: we \: know \: the \: formula \: \rightarrow }} \\ \\ \bf{ an = a + (n - 1) \times d} \\ \\ a = 12, \: d = a2 - a1 = 18 - 12 = 6 \\and \: last \: term = 96 \\ \\ \tiny \underline{\bf{we \: have \: find \:how \: many \: two \: digits \: numbers \: are \: divisible \: of \: 6}} \\ \\ > > 96 = 12 + (n - 1) \times 6 \\ \\ > > \: 96 = 12 + 6n - 6 \\ \\ > > \: 96 = 6n + 6 \\ \\ > > \: 6n = 96 - 6 \\ \\ > > \: 6n = 90 \\ \\ > > \: n = \frac{90}{6} = 15 \\ \\ \blue{\tiny \bf{ so, \: there \: are \: 15 \: numbers \: those \: which \: are \: divisible \: of \: 6}}$