MAH MBA CET Number Systems Questions

Prime numbers from 0 to 100 are 2,3,5,7,11,........
The multiplication will be very complicated. So, the question requires you to find number of zeros at the end.
Product = 2*3*5*7.....
No. of zeros at end = highest power of 10, which is 1 in this case(since 2,5 occur only once each and no higher powers of 2 and 5 occur).
So, the answer is option D as it has only 1 zero at the end.

17*17*18*18*19*19 = $$2^2\times\ 3^4\times\ 17^2\times\ 19^2$$
971 is not divisible by 2 or 3 or 17 or 19
173,23,5813 are also not divisible by 2 or 3 or 17 or 19
646=2*19*17. So, it can divide the given no.
Option C is the answer.

R($$\ \frac{\ 1719*1721*1723*1725*1727}{18}$$) = R($$\frac{9\times\ 11\times\ 13\times\ 15\times\ 17\ }{18}$$) = 9

Total even integers between 100 and 200 = $$\frac{200-100}{2}\ \ +\ \ 1\ =\ 51$$
Even integers divisible by 7 = Number of multiple of 14.
There are 7 integers which are of form 14k(112,126,...) between 100-200
Even integers divisible by 9 = Number of multiple of 18.
There are 6 integers which are of form 18k(108,126,...) between 100-200
Even integers divisible by both 7 and 9 = No. of multiple of LCM(14,18) = No. of multiple of 126k = 1
So, Required answer = 51-(7+6)+1 = 39.
Option C is the answer.

Let the no. be x
LCM(x,990) = 6930
6930=$$2\times3^2\times5\times7\times11$$
990=$$2\times3^2\times5\times11$$
Factor of 7 is missing in 990, so x has to be multiple of 7. 
x=7a
GCD(x,550)=110
x=110b
x=7a=110b. So, x=770k
Least value of x is 770.
Sum of digits of 770=7+7+0=14
Option B is the answer.    

Unit digit of $$6374^{1793}\times625^{317}\times341^{491}\ $$ = Unit digit of $$4^{1793}\times5^{317}\times1^{491}\ $$ = unit digit of 4*5*1 = 0
Option A is the answer.

To get the last digit, we could simply find the product of last digits of all the numbers i.e. 1*2*3*4*5*6*7*8*9.
As, there is 5 and 2 in the multiplication, 10 becomes a factor of the product and thus last digit will surely be 0.

1420 can be factorized as : $$2^2\times\ 5\times\ 71$$.
Total number of divisors = (2+1)(1+1)(1+1) = 3*2*2 = 12.

148 = 4 $$\times$$ 37

185 = 5 $$\times$$ 37

LCM (148, 185) = 4 $$\times$$ 5 $$\times$$ 37

$$\Rightarrow$$ LCM(148, 185) = 740

Hence, option A is the correct choice.

A number is divisible by 3 if the sum of its digits is divisible by 3.

Option A: 2345678 $$\Rightarrow$$ Sum of digits = 35 $$\Rightarrow$$ Not divisible by 3

Option B: 2876423 $$\Rightarrow$$ Sum of digits = 32 $$\Rightarrow$$ Not divisible by 3

Option C: 4006020 $$\Rightarrow$$ Sum of digits = 12 $$\Rightarrow$$ Divisible by 3

Option D: 9566003 $$\Rightarrow$$ Sum of digits = 29 $$\Rightarrow$$ Not divisible by 3

Option E: 3456785 $$\Rightarrow$$ Sum of digits = 38 $$\Rightarrow$$ Not divisible by 3

Hence, option C is the correct choice.

Three numbers are in the ratio 3 : 4 : 5

Let the numbers be 3k, 4k and 5k

LCM(3k, 4k , 5k) = 60k

60k = 2400 $$\Rightarrow$$ k = 40

The numbers are 120, 160, 200

$$\therefore$$ HCF(120, 160, 200) = 40

Hence, option C is the correct choice

The number of zeroes at the end of 1090! will depend on the power of 5 in the prime factorization of 1090!.
Number of powers of 5 can be calculated as 1090/5 + 218/5 + 43/5 + 8/5 = 218 + 43 + 8 + 1 = 270.
Thus, the number of zeroes at the end of 1090! are 270.

The remainder when 9 is divided by 8 = 1.
So, the remainder when $$9^{100}$$ is divided by 8 = $$1^{100}\ =\ 1$$ i.e. Option A.

$$128^{1000}$$ can be written as $$2^{7000}$$ and the divisor i.e. 153 can be broken down into 9*17. 
$$2^{7000}$$ can be written as $$8^{2333}\times\ 2$$. 8 leaves -1 as remainder when divided by 9. So, $$8^{2333}\times\ 2$$ will leave $$-1^{2333}\times\ 2$$ as remainder i.e. -2 when divided by 9. So, the remainder when the number is divided by 9 is 9-2 = 7.

Similarly, $$2^{7000}$$ can be written as $$16^{1750}$$ and 16 leaves -1 remainder when divided by 17 so $$16^{1750}$$ will also leave $$-1^{1750\ }$$ i.e. 1 remainder when divided by 17.

Now, the remainder can be written as: N = 9a + 7 = 17b + 1.
17b - 9a = 6. Possible values for a and b are 5 and 3 rsepctively.
Thus, N = 52 and the number can be written as 153K + 52 thus leaving 52 as remainder when divided by 153. 

For a number to be divisible by 99, it has to be divisible by both 9 and 11.
Only 114345 i.e. Option D is divisible by both 9 and 11 and thus it is the correct answer.

If a number is divisible by each of 15, 21 and 28, then it should also be divisible by their LCM i.e. 420.
10000 when divided by 420 leaves remainder 340.
It means that the largest number which is divisible by 420 and is lower than 10000 is 10000 - 340 = 9660.
Thus, the largest four-digit number exactly divisible by 15,21 and 28 is 9660.

Numbers having 4 factors are either of the form $$a^3\ or\ ab$$ where a and b are prime factors.
Numbers of the form $$a^3$$ which are less than 100 are: 8 and 27 = 2 numbers.
To find the numbers of the form ab, we will assume values for "a".
Case 1: a = 2. For ab to be lower than 100, "b" can take values 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47 = 14 numbers.
Case 2: a = 3. For ab to be lower than 100, "b" can take values 5, 7, 11, 13, 17, 19, 23, 29 and 31 = 9 numbers.
Case 3: a = 5. For ab to be lower than 100, "b" can take values 7, 11, 13, 17 and 19 = 5 numbers.
Case 4: a = 7. For ab to be lower than 100, "b" can take values 11 and 13 = 2 numbers.
No other cases are possible.
So total cases possible for numbers lower than 100 having 4 factors are 2 + 14 + 9 + 5 + 2 = 32.

Let us assume that largest number be A and the common remainder to be B.
64 = Ax + B
136 = Ay + B
238 = Az + B where x, y and z are quotients.
Subtracting 1st equation from 2nd, we get: 72 = A(y - x). So, A must be a factor of 72.
Subtracting 2nd equation from 3rd, we get: 102 = A(z - y). So, A must be a factor of 102.
Subtracting 1st from 3rd, we get: 174 = A(z - x). So, A must be a factor of 174.
So, the HCF of 72, 102 and 174 is 6 which means that the maximum value that A can take is 6.

For any number to be divisible by 11, the difference between the sum of digits at odd and even positions should be a multiple of 11.
Sum of odd position digits = A + 8
sum of even position digits = 3 + 1 = 4.
Their difference i.e. A+8 - 4 = A+4 should be a multiple of 11. So, the smallest value that A can take for which A+4 is a multiple of 11 is 7 i.e. Option D.

The maximum value of n will be equal to the number of powers of 5 in the expansion of 146!.
This will be equal to whole number values of: $$\frac{146}{5}+\frac{29}{5}+\frac{5}{5}=29\ +\ 5\ +\ 1\ =\ 35$$.
So, the maximum value that n can take is 35.

3556 can be factorized as $$2^2\times\ 7\times\ 127$$ and,
3444 can be factorized as $$2^2\times3\times\ 7\times\ 41$$.
The HCF will be $$2^2\times7\ =\ 28$$.

We will have to find the HCF of 4.5, 9.9 and 16.2 metres.
Converting them into cm to get 450, 990 and 1620 cm. HCF of these 3 numbers is 90 cm.
So, the length of the plank which can be used to measure exactly the lengths 4 m 50 cm, 9 m 90 cm, and 16 m 20 cm in the least time is 90 cm.

Let the number be A.
As 70 leaves 1 remainder when divided by A, it could be written as: 70 = Ax + 1 where x is the quotient.
Similarly, 50 = Ay + 4.
Subtracting the equations, we get: 20 = A(x - y) - 3 ==> A(x-y) = 23.
23 can be expressed as a product of 2 numbers in only 1 way as 23 is a prime number.
So, A can either be 23 or 1 and the greatest value that A can take is 23.

For any number to be divisible by 9, the sum of the digits of the number has to be divisible by 9.
Sum of digits of 381A is 12+A. The smallest value of A to make 12+A a multiple of 9 is 6.
Thus, the smallest natural number value of A is 6.