Test Level 3: Number System - 2 - CAT MCQ

1² = 1, 11² = 121, 111² = 12321, 1111² = 1234321 and 11111² = 123454321
⇒ 11111111² = 123456787654321
The sum of the digits of given number is 64.

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

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.

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 4^th 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.

 (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

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.

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.

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

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

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.

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.

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.

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

Since 300 < abc < 400
^⇒ 100a + 10b + c = a³ + b³ + c³
Put a = 3 and rearrange:
300 + 10b + c = 27 + b³ + c^3
⇒ (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.

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.

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

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.

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.

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.

A factor pair search of 2430 would give you the answer as 45 × 54.
Hence, 45 is the correct answer.