Question

how many two digit numbers are multiples of 4​

Answer 1

Answer:

the last two digits possible for thenumber to be a multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6 differentnumbers possible. So, with this set of 5digits we can have 36 different numbers.

Step-by-step explanation:

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Answer 2

Answer:

There are 22 two-digit terms in the sequence which is a multiple of 4.

Step-by-step explanation:

As we all learn that the first two-digit number exists as 10 and the last two-digit number is 99.

But none of these exists divided by 4.

The first two-digit number which exists with a multiple of 4 is 12 and the second, a two-digit number that stands the multiple of 4 is 16 and the last two-digit number which exists with a multiple of 4 is $96 .$

So the set of two digits numbers that are multiples of 4 stands$=\{12,16, \ldots \ldots \ldots \ldots ., 96\} .$

Now as we see that the above series form an arithmetic progression (A.P), whose first term (a) is 12, common difference

$(\mathrm{d})=(16-12)=(20-16)=4$

and last term $\left(a_{n}\right)=96$

Now as we understand that in an A.P the last term $\left(a_{n}\right)$formula is presented as,

$\Rightarrow a_{n}=a+(n-1) d$, where

$\mathrm{n}$ is the number of terms.

Now substitute the values we hold,

$\Rightarrow 96=12+(n-1) 4$

Now simplify this we have,

$&\Rightarrow 96-12=84=4(n-1)$

$&\Rightarrow \frac{84}{4}=21=(n-1)$

$&\Rightarrow n=21+1=22$

So there are 22 two-digit terms in the sequence which is a multiple of 4.

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