Free Numbers Set II - 01 Practice Test - CAT 

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.

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 $5=340$

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

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

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

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

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)

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 $9=1+3+7=(4+7{)}_9=12\Rightarrow \frac{1+2}{8}{\mid }_R=3$

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.

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)

$\frac{x}{2}$ is a perfect square $\Rightarrow \frac{x}{2}=a^2.\frac{x}{3}$ is a perfect cube $\Rightarrow \frac{x}{3}=b^3.\frac{x}{2}\times \frac{x}{3}=a^2\times b^3\Rightarrow x=\sqrt{6a^2{b}^3}=ab\sqrt{6b}.$
For x to be a natural number; 6b has to be a perfect square $\Rightarrow b=6\Rightarrow x=2^3\times 3^4\Rightarrow$ number of divisors $=20$

We know that Number = Quotient $\times$ Divisor + Remainder

If Divisor (a-2) equals 0, then Number = Remainder. This is the property we are going to use

Make the divisor=0; i.e. put a=2, Then $a^3+3a^2-am+4=m$ where a=2

Thus, 3m= 24 and m=8.

$(11100{)}_2=0+0+2^2+2^3+2^4=28$

$(1001{)}_x=x^0+x^3=1+x^3$

$1+x^3=28=27+1=3^3+1,x=3.$

If a number is divisible by 72, it is divisible by 8 and 9. To be divisible by 8, the number formed by the last three digits should be divisible by 8. Therefore, 31Y should be divisible by 8$\Rightarrow$ Y = 2. To be divisible by 9, the sum of the digits of the number should be divisible by 9$\Rightarrow$ X = 3. Therefore, X-Y = 1

Use Chinese remainder theorem,

Let the number be N. N is of the form 9A+2 (i.e. N divided by 9 gives a remainder 2)= 10B+3 (i.e. N divided by 10 gives a remainder 3) =11C+5 (i.e. N divided by 11 gives a remainder of 5)

We will first find the number which is of the form 9A+2=10B+3.--------(1)

Here for B=8, A=9, which are the first integer solutions.

Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3.

The numbers will fall in an AP with the first digit as 83 and common difference 90 (9*10). The general form of the AP is 83+90k when k=0,1,2.....

Now to include the 3${}^{rd}$ condition(i.e divisor 11 and remainder 5) we can write 83+90k=11C+5 -------(2)

Find the first integer value of K which satisfies the equation such that C is also an integer.

Here for K = 5, C = 48, which are the first integer solutions. Substituting this in equation

(2) we get the value 90 (5) + 83= 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5.

Shortcut:- You may be tempted to mark option (a) after a first glimpse at the options.

Since the question asks for the least number, you should be careful while answering.

You can still back calculate using answer option (a)

option (a) 1523 satisfies the condition. Hence 1523-990 = 533 has to be the first such

number. Answer is option (e)

Also note that, the first number has to be $<$ 990.