After 10!, the factorials end with 0 as the last two digits. We thusneed to find the last two digits from 4! To 9! Only

4!+5!+6!....9!= 24+120 + 720 + 5040 + 40320 + 362880 = ___04

Note that 71! has 16 zeroes at its end.(You can find this out by finding the highest power of 5 in 71!)

So does 70!.

66! To 69! Have 15 zeroes at its end. Multiplication of these itself amounts to $>202$

Hence the ${202}^{nd}$ digit is 0

To find the highest power of a number in a factorial

a)Highest power of a prime number in a factorial:

To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

Eg) Find the highest power of 100!

Solution:

$\frac{100}{5}=20;\frac{20}{5}=4;$

Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24

b)Highest number of a composite number in factorial

1)Factorize the number into primes.

2)Find the highest power of all the prime numbers in that factorial using the previous method.

3)Take the least power.

32540 using the divisibility test for 13 ( triplets at odd- triplets at even places ) can be written as

540-32= 508. 508 when divided by 13 can be represented as 13k + 1 (i.e. when the number is divided by 13, the remainder is 1)

Similarly

42400= 400-42= 358 = 13k + 7

3233= 233-3=230= 13k + 9

(13k + 1)( 13k + 7)(13k + 9) will give the non-divisible part as $(9\times 7\times 1)=63$

therefore 63+222= remainder of 285 when divided by 13= -1 or 12

Write all the numbers in base $10$ and then arrive at the answer

$127$ in base $8=7\times 1+2\times 8+1\times 8^2=87$ in base $10$

$12$ in base $6=2\times 1+1\times 6=8$ in base $10$

Adding both $=95$ in base $10$

Expressing in base $5=340$

Following are the steps while calculating the answer

4 digit numbers divisible by $11=\frac{8999}{11}=819,4$ digit numbers divisible by $13=\frac{8999}{13}=692$ or $693$

Number of integers common to 11 and 13 (as questionis 11 or 13) $=\frac{8999}{(13\times 11)}=62$ or $63$

To subtract numbers which are common to $11\&17;13\&17;11,13\&17$

$11\&17=\frac{8999}{(11\times 17)}=47$ or $48$ terms

$13\&17=\frac{8999}{(13\times 17)}=40$ or $41$

$11,13,17=\frac{8999}{(11\times 13\times 17)}=3$ or $4$ terms

Total ( you can approximate as the answers are far apart)= 819+693-63-48-41+4= 1364

Closest answer is option (2)

Note: This is based on the concept of Euler’s number

A number N can be written as $a^m{b}^n$. In this case, $N=900=2^2\times 3^2\times 5^2$

The question can be rephrased as

The number of natural numbers which are less than 900 and relatively prime to it is

$N=[1-\left(\frac{1}{a}\right)][1-\left(\frac{1}{b}\right)][1-\left(\frac{1}{c}\right)]$

$N=[1-\left(\frac{1}{2}\right)][1-\left(\frac{1}{3}\right)][1-\left(\frac{1}{5}\right)]=240$

10 factors implies that the number can be of the form

1) $a^9$ ( Least odd number will be $3^9$)

2) $a{b}^4$ (Least odd number will be $3^4\ast 5=405$)

Therefore, as the least number is 405, there are no two digit numbers of the specified form. The answer is 0.

If a number N can be factorized as $N=a^m\ast b^n$(a, b are the prime factors)

Then number of factors= (m+1)(n+1)

The first power of 10 for which a division by 13 gives a remainder -1 is x=3. the next power of 10 will be ${10}^9$ . hence it falls in an AP with first term as 3 and last term as 99 with a common difference of 6

number of terms $=\frac{(99-3)}{6}+1=\frac{96}{6}+1=16+1=17$ terms.

In base 9, divisibility check for 8 is sum of the digits(digit sum) =Digit sum of $N=(25\ast 3+25\ast 2{)}_9=$

First convert the numbers to base 10

$25$ in base $10=23,$ Thus the question changes to $(23\ast 3+23\ast 2{)}_{10}={115}_{10}$

Converting it back to base 9 = 137, Now sum of digits in base $9=1+3+7=(4+7{)}_9=12\Rightarrow \frac{1+2}{8}{\mid }_R=3$