This EduRev document offers 20 Multiple Choice Questions (MCQs) from the topic Number System (Level - 3). These questions are of Level - 3 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 3: Number System - 2
Try yourself:If N = (11111111)2, then what is the sum of the digits of N?
Explanation
12 = 1, 112 = 121, 1112 = 12321, 11112 = 1234321 and 111112 = 123454321
⇒ 111111112 = 123456787654321
The sum of the digits of given number is 64.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:A is the smallest integer which when multiplied by 3 gives a number made of 5's only. What is the value of C3, if sum of the digits of A is B and sum of the digits of B is C?
Explanation
Smallest number made of 5's and divisible by 3 is 555.
Thus, A = 555/3 = 185
B = 1 + 8 + 5 = 14
C = 1 + 4 = 5
Thus, value of C3 = 53 = 125
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:ab and cd are 2 two-digit natural numbers. Given that 4b + a = 13k1 and 5d - c = 17k2, where k1 and k2 are natural numbers. Find the largest number that will always divide the product of ab and cd.
Explanation
ab × cd = (10a + b)(10c + d)
= (40b + 10a - 39b)(51d - (50d - 10c))
= (10(4b + a) - 39b)(51d - 10(5d - c))
= (10 × 13k1 - 39b)(51d - 10 × 17k2)
= 13 × 17(10k1 - 3b)(3d - 10k2)
= 221(10k1 - 3b)(3d - 10k2)
Hence, the largest number that will divide the product ab and cd is 221.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:Find the last digit of the following expression.
(15)256 + (19)138 + (32)97
Explanation
Unit digit of (15)256 = 5.
As any number with 5 at the units place will always have a 5 at its units place, irrespective of the positive natural number power.
Now, 91 = 9, 92 = 81, 93 = 729, 94 = 6561
So, basically, a number with 9 at its units place raised to an even number, will have 1 at its units place and a number with 9 at its units place raised to an odd number, will have 9 at its units place.
Thus, the units digit of 19138 is a 1.
Now, 21 = 2, 22 = 4, 23 = 8, 24 = 16 and 25 = 32
So, basically after every 4th power, the units digit repeats itself.
Now, 97 = 96 + 1 = 24 x 4 + 1
Thus, we have 24 sets of 4 and then 1 more.
So, the number should have 2 at the units place.
Thus, the digit at the units place is 5 + 1 + 2 = 8
Hence, option 1 is correct.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:Find the value of a in the identity (11100)2 = (1001)a.
Explanation
(11100)2 = (28)10
(1001)a = 1 × a3 + 0 × a2 + 0 × a + 1 = a3 + 1 = (1 + a3)
According to the question, (28)10 = (a3 + 1)10
⇒ a3 = 27
⇒ a = 3
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:A number when divided by 296 leaves the remainder 75. If the same number is divided by 37, then what will be the remainder?
Explanation
Let x be the number.
Given: x = 296n + 75 [where n is the quotient]
x = 37(8n + 2) + 1
Hence, when x is divided by 37, it gives 1 as the remainder.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:If N = (7p + 4)(5q)(23) is a perfect cube, where p and q are positive integers, then find the smallest possible value of p + q.
Explanation
In order for N to be a perfect cube, each prime factor of N must have the exponent that is divisible by 3.
Since p and q must be positive integers, the smallest value of q and p is 3 and 2, respectively.
Thus, the smallest value of p + q is 5.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:There are four prime numbers written in ascending order of magnitude. The product of the first three is 385 and that of the last three is 1001. Find the first number.
Explanation
The best way is to factorise the number i.e 385 = 5 × 7 × 11
1001 = 7 × 11 × 13
So, the four prime numbers are 5, 7, 11 and 13.
Hence, first prime number = 5
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:What is the greatest positive power of 5 that exactly divides 30!?
Explanation
The total number of multiples of 5 in 30! = 6 (5, 10, ….., 30)
The total number of multiples of 5 to the power 2 = 25 in 30! = 1 (25 only)
Further powers of 5 cannot be these as 5 to the power 3 = 125 > 30
Therefore, the greatest power of 5 that divides 30! exactly = 6 + 1 = 7
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:How many multiples of 32 are perfect squares, less than 104?
Explanation
The least multiple of 32, which is a perfect square is 32 × 2 = 64.
So, all the numbers of the form 64k (where k is a perfect square) are perfect squares.
i.e. 64 × 12, 64 × 22, 64 × 32, 64 × 42, … 64 × 122
If we take 64 × 132;
64 × 169 > 104
So, total 12 numbers are there.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:Let A, B and C be positive integers satisfying A < B < C and A + B + C = X. What is the smallest value of X that does not determine A, B and C uniquely?
Explanation
We have to check the options one by one.
For X = 9, A + B + C = 9, we have more than one solution, i.e. (1, 2, 6), (1, 3, 5), etc. So, X = 9 does not determine A, B and C uniquely.
For X = 6, we have a unique solution, i.e. (1, 2, 3).
For X = 7, we have a unique solution again, (1, 2, 4).
For X = 8, we have more than one solution, i.e. (1, 2, 5); (1, 3, 4).
Hence, 9 and 8 are the two values that do not determine A, B and C uniquely. Of the two, 8 is the required smallest value.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:If, in a number system, 40132 corresponds to 2542 in decimals, find the base of the number system.
Explanation
Since 4 is one of the characters in the given number, the base for that number system must be > 5.
Assume that the base is 5. Then, (4 × 54) + (0 × 53) + (1 × 52) + (3 × 51) + (2 × 50) = 2542
So, by estimation, we can say that the base of the number system is 5.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:Let D be a decimal of the form D = 0.abcdabcdabcd …, where the digits a, b, c and d are integers lying between 0 and 9. At most three of these digits are zero. By what number should D be multiplied so that the result is a natural number?
Explanation
D = 0.abcdabcdabcd...
10000D = abcd.abcdabcd...
10000D - D = (abcd.abcdabcd...) - (0.abcdabcdabcd...)
9999D = abcd
D = abcd/9999
From the options, 49995 is the only multiple of 9999.
49995 = 9999 × 5
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:If abc is a three-digit whole number, such that abc = a3 + b3 + c3. [300 < abc < 400], find the sum of a, b and c.
Explanation
Since 300 < abc < 400
⇒ 100a + 10b + c = a3 + b3 + c3
Put a = 3 and rearrange:
300 + 10b + c = 27 + b3 + c3
⇒ (b3 - 10b) + (c3 - c) = 273
Now, put b = 6 and b = 7 (as b and c are less than equal to 9, and 63 and 73 are the nearest cubes of 273)
For b = 6, we get c3 - c = 117, and no value of c satisfies it.
For b = 7, we get c3 - c = 0, and two values of c, i.e. 0 and 1 satisfy it.
Thus, abc = 370 or 371
Hence, abc cannot be uniquely defined. So, option 4 option is the answer.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:A number consists of three consecutive digits in order, the one in units place being the greatest of the three. The number formed by reversing the digits exceeds the original number by 22 times the sum of the digits. Find the number.
Explanation
Let the digits at hundred's place be x.
Hence the number will have the digits as x, x + 1 and x + 2.
Number will be 100x + 10x + 10 + x + 2 = 111x + 12
Number obtained after reversing the digits = 100x + 200 + 10x + 10 + x = 111x + 210
Sum of the digits = x + x + 1 + x + 2 = 3x + 3
According to the question,
⇒ 111x + 210 = 111x + 12 + 22(3x + 3)
⇒ 210 = 12 + 66x + 66
⇒ 66x = 132
⇒ x = 2
Hence, the number will be 234.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:If the first 100 natural numbers are written side by side to make a big number, how many digits of this number will be even?
Explanation
Single-digit even numbers are 4 in number (2, 4, 6 and 8).
In two-digit numbers, i.e. from 10 to 99,
0 comes at unit place in 10, 20, 30, 40 ....90 = 9 times
2 comes at unit place in 12, 22, 32, 42 ....92 = 9 times
2 comes at tens place in 20, 21, 22, 23, 24.....29 = 10 times
Thus, 2 comes from 10 to 99 = 9 + 10 = 19 times
Similarly 4, 6 and 8 come 19 times each.
In 100, there are 2 zeros.
Total number of even digits = 4 + 9 + 19 + 19 + 19 + 19 + 2 = 91
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:For how many integers n, with 10 ≤ n ≤ 100, is n2 + n - 90 divisible by 17?
Explanation
Since n2 + n - 90 = (n - 9)(n + 10)
17 divides n2 + n - 90 if and only if 17 divides n - 9 or n + 10.
For n + 10 to be divisible by 17, n can be 24, 41, 58, 75 or 92.
For n - 9 to be divisible by 17, n can be 26, 43, 60, 77 or 94.
So, n has 10 values.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:How many three-digit numbers do not change even if their digits are reversed?
Explanation
A three-digit number doesn't change even if the digits are reversed when the first and last digits of the number are same.
Since the first digit cannot be zero, it can be anything from 1 to 9 and the second digit can be anything from 0 to 9.
So, the first and last digits can be filled in 9 ways and the middle can be filled in 10 ways.
So, total such numbers are 9 × 10 = 90.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:Find the lower of the two successive natural numbers if the square of the sum of those numbers exceeds the sum of their squares by 112.
Explanation
Again, spotting this with options is quite easy as we can see that 72 + 82 = 113 and that is 112 less than the value of (7 + 8)2 = 225.
Without options here you can think of a2 + b2 + 112 = (a + b)2 → 2ab = 112 or ab = 56.
Since, the numbers are consecutive, sifting through the factor pairs of 56 we can see the numbers as 7 & 8, respectively.
Report a problem
Question for Practice Questions Level 3: Number System - 2
Try yourself:The product of a two-digit number by a number consisting of the same digits written in the reverse order is equal to 2430. Find the lower number.
Explanation
A factor pair search of 2430 would give you the answer as 45 × 54.
Hence, 45 is the correct answer.
Report a problem