CAT Remainders Questions

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 - 60 - 12 - 7 = 41

At $$x = 0, 2^x = 1$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution. 

At $$x = 1, 2^x = 2$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution. 

At $$x = 2, 2^x = 4$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution. 

At $$x = 3, 2^x = 8$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution. 

At $$x = 4, 2^x = 16$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution. 

At $$x = 5, 2^x = 32$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution. 

At $$x = 6, 2^x = 64$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution. 

At $$x = 7, 2^x = 128$$ which is in the given range [0.25, 200]

$$2^x + 2$$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution. 

At $$x = 8, 2^x = 256$$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.

Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values. 

According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!.  So p = 11! - 1! = 11! - 1.  p + 2 = 11! + 1 .So when it is  divided by 11! gives a remainder of 1. Hence, option 4.

We know that x = $$16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$$

= $$(16 + 19)(16^2 - 16 * 19 + 19^2) + (17 + 18)(17^2 - 17 * 18 + 18^2)$$ = 35 × odd + 35 × odd = 35 × even = 35 × (2k)

=> x = 70k

=> Remainder when divided by 70 is 0.

The remainder when $$15^{23}$$ is divided by 19 equals $$(-4)^{23}$$
The remainder when $$23^{23}$$ is divided by 19 equals $$4^{23}$$
So, the sum of the two equals$$(-4)^{23}+(4)^{23}=0$$

Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 - (7+6-1) = 39 

There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

positive integers n in the range $$12 \leq n \leq 40$$ such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.

$$2^4 = 16 = -1$$ (mod $$17$$)
So, $$2^{256} = (-1)^{64} $$(mod $$17$$)
$$= 1$$ (mod $$17$$)
Hence, the answer is 1. Option a).

Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K 

Let k = 1; the number becomes 53

If it is divided by 84, the remainder is 53.

Option d) is the correct answer.

Alternative Solution.

Consider only for 3 and 4 and the remainders are 2 and 1 respectively.

So 5 is the first number to satisfy both the conditions. The number will be of the form 12k+5. Put different integral values of k to find whether it will leave remainder 5 when divided by 7. So the first number to satisfy such condition is 48x4+5= 53

Consider n=1 we have $$7^{6} - 6^{6}$$ which is = $$(7^{3} + 6^{3})(7^{3} - 6^{3})$$ = 13 * 127 * 43 which is divisible by all the 3 options.

Option d) is the correct answer.

We know that a=$$b^2-b$$.

So$$a^2-a$$ = b($$b^3-2b^2-b+2$$) . = (b - 2)(b - 1)( b)(b + 1)

The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)

In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).

Thus, for any value of b >=4, $$a^2-4$$ would be divisible by 3 x 2 x 4 = 24.

Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $$^4C_3 $$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

The difference of the numbers = 34041 - 32506 = 1535

The number that divides both these numbers must be a factor of 1535.

307 is the only 3 digit integer that divides 1535.

$$55^3 + 17^3 - 72^3$$ = $$(55-72)k + 17^3$$. This is divisible by 17

Remainder when $$55^3$$ is divided by 3 = 1

Remainder when $$17^3$$ is divided by 3 = -1

Remainder when $$72^3$$ is divided by 3 = 0

So, $$55^3 + 17^3 - 72^3$$ is divisible by 3

So, the answer is d) 3 and 17

$$7^3$$ = 343

$$7^{84}$$ = $$(7^3)^{28}$$ = $$343^{28}$$

$$343^{28}$$ mod 342 = $$1^{28}$$ mod 342 = 1

Let's say N is our number
N = (899K + 63) or N = ($$29 \times 31$$K) + 63
So when it is divided by 29, remainder will be $$\frac{63}{29}$$ = 5

For a number to be divisible by 8, last 3 digits must be divisible by 8.

Last 3 digits of this number are 354.

354 mod 8 = 2

Hence, 2 is the remainder.

Let's say m=5k and n=5t
So m-n = 5(k-t) will be divisible by 5.
$$m^2 - n^2 = 25(k^2 - t^2)$$ will be divisible by 5.
$$m+n = 5(k+t)$$ will be divisible by 5 but not necessarily with 10.

As 55 and 124 don't have any common factor, and n has to be a minimum integer, Hence, it should be 124 only. So that given equation won't have a remainder.

let's say k+4 = 7m
k = 7m-4
Now for k+2n or 7m+(2n-4) is also  multiple of 7.
or 2n-4 should be a multiple of 7 
So 2n-4 = 7p
or 2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer )

Let the Prime number be 6n+1.

So $$(p^2 - 1)$$ = 6n(6n+2) = 12n(3n+1)

For any value of n , n(3n+1) will have a factor of 2

Hence given equation will be always be divisible by 24

$$2^{60}$$ or $$4^{30}$$ when divided by 5
So according to remainder theorem 
remainder will be $$(-1)^{30}$$ = 1.