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# The remainder when $1!+2!+3!+.....95!$ is divided by 15 is - (a) 3(b) 5(c) 8(d) 12 Verified
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Hint: Here in this question, we have to find the reminder of the given factorial sum. For this, we know that the value of factorials after 5! will leave the remainder zero when it is divided by 20. So, for calculating the remainder, we will calculate the remainder before that number.

So, first we will calculate the factorial till 5!. So, for this, the factorials will be,
\begin{align} & 1!=1,2!=2\times 1,3!=3\times 2\times 1=6, \\ & 4!=4\times 3\times 2\times 1=24 \\ \end{align}
And the factorial after 5! will have the remainder zero. So, now adding all the factorial values from 1 to 5, we get $\Rightarrow \dfrac{1!+2!+3!+4!}{15}$
And on substituting the values, we had obtained above, we will get the equations as $\Rightarrow \dfrac{1+2+6+24}{15}$
Now, adding the numerator of the above fraction, we get the fractions as $\Rightarrow \dfrac{33}{15}$
And now on dividing it, we get the remainder as 3.
Therefore, the remainder for the factorial,
$1!+2!+3!+.....95!$
Note that $5!=1\times 2\times 3\times 4\times 5$ is divisible by 15, and hence for any number, $n\ge 5$, we have $n!=1.2.3.4.5.....n$ is divided by 15 and these are divisible.
Therefore, the remainder when $1!+2!+3!+.....95!$ is divided by 15 the same as the remainder when 15 divides $1!+2!+3!+4!$.
Now, $1!+2!+3!+4!=1+2+6+24=33$ is divided by 15, then the remainder is 3.

So, the correct answer is “Option a”.

Note: 20 can be factored as a product of 5 and 4. So, the least value of factorial which contains both 4 and 5 is 5!. So, 5! And any factorial greater than always contains both 5 and 4. So, the remainder is going to be zero for any x!.
Whether x is either 5 or greater than 5. For solving this type of question we should know how to calculate the factorial of any number. And also if we memorize the properties then we don’t have to solve and add fractional value.