Test Level 2: Number System - 2 - CAT MCQ

1! = 1 (Units digit = 1)
2! = 2 × 1 = 2 (Units digit = 2)
3! = 3 × 2 × 1 = 6 (Units digit = 6)
4! = 4 × 3 × 2 × 1 = 24 (Units digit = 4)
5! = 5 × 4 × 3 × 2 × 1 = 120 (Units digit = 0)
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 (Units digit = 0)
So, we can see that the units digit of all the factorials from 5! onwards is 0.
So, the units digit of the given sum depends upon the sum of the units digits of 1!, 2!, 3! and 4!, which is 3 (6 + 4 + 2 + 1).

385 = 5 × 7 × 11
Hence, 2-digit factors are 11, 35, 55 and 77

Co-prime means, the integers which have their HCF as 1, they don`t have any common factor other than 1.
So, the numbers must be 1 and 8.
∴ Sum = 1 + 8 = 9

Let (u, v, w) be (4, 1, 3), respectively.
Start testing from the last choice.
(5): uv + w = 4 + 3 = 7, which is odd
(4): (v + w)² - u² = 16 - 16 = 0,
So, we can stop here and choose (4) as the answer.
(You can independently verify that the other three choices are odd integers)

We cannot say anything about (1), (2) and (4).
But, if x and y are co-primes, then they do not have any common factors and hence, neither do their squares.

For the value abc to be the largest, one of them must be equal to 9. So, the possible values for (c, b, a) consistent with the equation a + b = c, are (9, 1, 8); (9, 2, 7); (9, 3, 6); (9, 4, 5); (9, 5, 4); (9, 6, 3); (9, 7, 2) and (9, 8, 1). Their products are 72, 126, 162, 180, 180, 162, 126 and 72, respectively. Among these, 180 is the largest.

Let us consider two sets of positive consecutive integers starting with an odd number and an even number, respectively i.e. (1, 2, 3) and (2, 3, 4).
I is true in both cases.
II is true in both cases.
III is not true in the first case.
So, (4) is the answer.

The last two digits of 25⁶³ are always 25.
Last two digits of 63²⁵ = (3^4 × 6)3 = (81)⁶ × 3 = 481 × 3 = 243 (An even number digit) 3
(Tens digit of any power of a number, whose units digit is odd and tens digit is even, is always even.)
Thus, the last two digits of 25⁶³ × 63²⁵ will be given by 25 × 43 (where a is even).
So, 75 is the required answer.

If we check, 99! ends with 22 zeros.

19 + 3 = 22
Likewise, 100! ends with 24 zeros.

20 + 4 = 24
So, none of the numbers ends with 23 zeros.

The number which has five factors should be of the form a⁴.
Such two-digit numbers are 2⁴ and 3⁴.

N = (15 × 1)(15 × 2)...(15 × 100)= 15¹⁰⁰ (1.2.3.4…100)
= 15¹⁰⁰ × 100!
= (3¹⁰⁰ × 5¹⁰⁰)(5²⁴ × 2⁹⁷ × K)
So, N will have total 97 zeros.

Let the number being divided be a.
Let the divisor be y and the quotient be z. Since a = zy + 23 (given), 2a = 2zy + 46.
Since the remainder is only 11 when 2a is divided by y, the remainder is also divisible by y and its remainder will be 11. Therefore, 46 = y + 11. On solving, we get y = 35, which is the divisor.

If a and b are two numbers, then their Arithmetic mean is given by (a + b)/2 while their geometric mean is given by (ab)^0.5.
Using the options to meet the conditions we can see that for the numbers in the first option (6 and 54) the AM being 30, is 24 less than the larger number while the GM being 18, is 12 more than the smaller number.
Option (a) is correct.

Since 15n³, 6n² and 5n would all be divisible by n, the condition for the expression to not be divisible by n would be if x is not divisible by n. Option (c) is correct.

If the number is n, we will get that 22n + n = 23n is half the square of the number n.
Thus, we have
n² = 46 n → n = 46

If we assume the numbers as 16 and 4 based on 4:1 (in option a), the AM would be 10 and the GM = 8 a difference of 20% as stipulated in the question. Option (a) is correct.

990 = 11 × 32 × 2 × 5. The highest power of 990 which would divide 1090! would be the power of 11 available in 1090.
This is given by [1090/11] + [1090/121] = 99 + 9 = 108

For finding the highest power of 6 that divides 146!, we need to get the number of 3’s that would divide 146!.  
The same can be got by: [146/3] + [48/3] + [16/3] + [5/3] = 70.

Both 333⁵⁵⁵ and 555³³³ are divisible by 3,37 and 111.
Further, the sum of the two would be an even number and hence divisible by 2.
Thus, all the four options divide the given number.

The given conditions are satisfied for the number 24.