Test: Number System - 2 - CAT MCQ

Let the number be N.
In order to maximize the number of factors of N³, N² must be expressed as a product of as many prime factors as possible.
No. of factors of N² = 105 = 3 × 5 × 7
where a = 2  b = 4   c = 6
then power original number = (2 + 1) (4 + 1) (6 + 1)
∴  N² = (a)² (b)⁴ (c)⁶, where a, b and c are prime numbers.
∴  N³ = (a)³ (b)⁶ (c)⁹
Where N = a^p b^q c^r no = (p + 1) (q + 1) (r + 1)
Hence, the number of factors of N³
= (3 + 1) × (6 + 1) × (9 × 1) = 4 × 7 × 10 = 280.

Since 'ab' is a two - digit prime number and one of its digit is 3, it can assume any of the values among 13, 23, 31, 37,43, 53, 73 and 83.
As the absolute difference between the digits of the number is not a factor of 2, the number among the obtained numbers that satisfy the aforementioned condition are 37, 73 and 83. Hence, the number of values that 'ab' can assume is 3.

E = 3 + 8 + 15 + 24 + .... + 195 = 1 × 3 + 2 × 4 + 3 × 5 + 4 × 6 + .... + 13 × 15
∴ T_n = n (n + 2) and n = 13

= 7 × 11 × 13
Hence the sum of the prime factors of E
= 7 + 11 + 13 = 31.

LCM of 6, 4 and 3 = 12
Multiply by 12 of each number in power

⇒ 2¹⁴, 3⁹, 5⁸
So, ascending order is
5⁸ > 3⁹ > 2¹⁴  or 5^2/3 > 3^3/4 >2^7/6

Prime factorization of 44 is = 2 × 2 × 11
To express 44 as product of five distinct integers
So, we'll have to put 1 and –1.
The only possible way comes out to be:
44 = 2 × (–2) × 11 × 1 × (–1)
In this case the value of n would be 11 which is also the only possible value.

500! + 505! + 510! + 515!
= 500! (1 + 5k) (where k is a natural number)
So (5k + 1) won't be a multiple of 5.
Minimum value of n for which 500 ! is divisible by 5ⁿ = 1.
Maximum value of n for which 500! is divisible by 5ⁿ

= 100 + 20 + 4 = 124
Hence, there are 124 possible values of n.

Sum of a₀ + a₁ + .... + a₅₀ = 1 + 3 + .... + 101

Let first integer = (x – 1), then
Second integer = x; so..... on
According to question.
⇒ (x  – 1)² + x² + (x + 1)² = (x + 2)² + (x + 3)²
⇒ x² + 1 – 2x + x² + x² + 1 + 2x = x² + 4 + 2.x.2 + x² + 9 + 2.x.3
⇒ 3x² + 2 = x² + 4 + 4x + x² + 9 + 6x
⇒ 3x² + 2 = 2x² + 10x + 13
⇒ 3x² + 2 – 2x²  – 10x – 13 = 0
⇒ x² – 10x – 11 = 0
⇒ x² – 11x + x – 11 = 0
⇒ x (x – 11) + 1 (x – 11) = 0
⇒ (x + 1) (x – 11)
∴ x = –1 or 11

Here find the number of two–digit natural numbers such that unit digit is greater than their ten's digit.
In such natural numbers, we cannot take 0 or 1 in units place.
When we take 2 at unit's place, we obtain only 1 Such number is 12.
When we take 3 at unit's place, we obtain 2 such numbes are 13 and 23.
When we take 9 at unit's place, we obtain 8 such numbers.
So, number of such numbers is (1 + 2 + 3 + .... + 8) = 36
Hence, the required number has 72 digits.

In the given equation the right hand side contains the powers of 2 and 3 only; therefore the left hand side should contain the powers of 2 and 3 only.
Since (x – 1)(x – 2)(x – 3) is a product of three consecutive numbers, it will always contain either one or two multiples of 2 and one multiple of 3. Lets make two cases:
(1) If (x – 1) and (x – 3) are multiples of 2:
Let (x – 1) be equal to 2k; then (x – 3) is equal to 2(k + 1).
Now k and (k + 1) should both contain powers of 2 or 3 only. This is possible with k = 1, 2 or 3. Also if any of k or (k + 1) is a multiple of 3, (x – 2) will not be a multiple of 3 or 2. So again it will not satisfy.
(2) If (x – 2) is a multiple of 2:
Here (x – 1) and (x – 3) will both be odd, out of which only one will be a multiple of 3. Hence the other number will be a multiple of an odd number other than 3. So the equation can be satisfied only if that other odd number is 1. Hence taking one odd number as 1 we get 1 × 2 × 3 which is equal to 6.
Hence the equation is satisfied for x = 4 only.

One such number is 1 which has no factor other than itself.
If the number has only one prime factor i.e. it is of the form pa where p is a prime number and a is a natural number, then according to the question: (a + 1)² = p^a
This is possible only if a = 2 and p = 3. So the number is 9.
If the number has two prime factors then it would be of the type p^a × q^b, where p and q are two distinct prime numbers. Then according to the question:
(a + 1)² (b + 1)² = p^a × q^b
This is possible only if p and q are both 3. Since they are different, this is not a valid case. So there would no such case with two or more prime factors.
So there are only two such integers - 1 and 9.

The general term is of the form
(X + n(n + 2)) / (n + 2)
n(n + 2) is always divisible by (n + 2).
So we can say that n(n + 2) ± 1 would never be divisible by (n + 2).
If we put X = –1, the numerator and denominator of all the terms would be co-prime.

From option (a),

Using Binomial here

= 2 + 5 x + 9x² + 14x³ + .....
this is same series as given
Thus, option (a) is correct answer.

For the younger child ₹ 7.5 lakh should become 21 lakhs in 9 years.
Hence,
Amount = Principal + Simple Interest

Similarly, for the elder son, ₹ 7.5 lakh should become in 6 years.
Hence,  Amount = Principal + Simple Interest

Consider 3⁴ⁿ = (81)ⁿ = (1 + 80)ⁿ = 1 + 80q, q ∈ N
∴ ₃3⁴ⁿ = 380q + 1 = (81)^20q.3
Since the last digit of (81)^20q is 1, so the last digit of
₃3⁴ⁿ + 1 is 1 × 3 + 1 = 4

(x² – xy + y²) = (x + y)
Multiplying both sides by 2:
2(x²– xy + y²) = 2(x + y) ⇒ 2x² – 2xy + 2y² = 2(x + y)
(x – y)² + x² + y² = 2x + 2y

Integer solutions for (x, y):
Case 1: (0, 0) and (2, 2)
Case 2: (1, 2) and (1, 0)
Case 3: (2, 1) and (0, 1)
So there are six non-negative integer solutions.

a¹⁵ + a¹⁶ + a¹⁷ + .... + a⁵⁰|
Sum = a¹⁵ {1 + a + a²  + ..... a³⁵}

Since a is the root of equation x⁵ – 1 = 0,
a⁵ – 1 = 0 ⇒ a⁵ = 1

The cyclicity of each digit from 0 to 9 is a factor of 4.
Hence any digit raised to a power of the type 4k + 1 will always end in the same digit. Hence the answer is x¹³ + y¹³.

Putting n = 1, we get 5 + √19 whose integral part is 9.
Putting n = 2, we get 25 + 19 + 10√19 whose integral part is 25 + 19 + 43 which is again an odd number.
Now, through the options it can be judged that the greatest integer must always be an odd number.

Let P, Q and R be n, n + 2 and n + 4 respectively in ascending order.
According to the Question
3n = 2(n + 4) – 3 = 2n + 5
∴ n = 5
Thus, R = 5 + 4 = 9