Test: Number System - 1 - CAT MCQ

2⁴ × 3⁵ × 10⁴ = 2⁸ × 3⁵ × 5⁴
For perfect squares, we have to take only even powers of the prime factors of the number
The number of ways 2’s can be used is 5 i.e. 2⁰, 2², 2⁴, 2⁶, 2⁸
The number of ways 3’s can be used is 3 i.e. 3⁰, 3², 3⁴
The number of ways 5’s can be used is 3 i.e. 5⁰, 5², 5⁴
Therefore, the total number of factors which are perfect squares = 5 × 3 × 3 = 45
But this also includes the number 1.
Hence excluding 1, the required number is 45 – 1 = 44.

m² + 105 = n² ⇒ n² – m² = 105
⇒ (n – m) (n + m) = 105
Since m an d n are positive integers (n – m) < (n + m), then by splitting 105 in two factors, we get
⇒ (n – m) (n + m) = 1 × 105
For (n – m) = 1 and (n + m) = 105, (m, n) = (52, 53)
⇒ (n – m) (n + m) =  3 × 35
For (n – m) = 3 and (n + m) = 35 (m, n) = (16, 19)
⇒ (n – m) (n + m) =  5 × 21
For (n – m) = 5 and (n + m) = 21 (m, n) = (8, 13)
⇒ (n – m) (n + m) =  7 × 15
For (n – m) = 7 and (n + m) = 15, (m, n) = (4, 11)
Hence, there are four required pairs.
Shortcut approach :
Number of pairs = 
105 = 3 × 5 × 7
Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105
Number of factors of 105 = 8
Hence, required number of pairs = 8 / 2 = 4

Let the number be ABCDEF, where A, B, C, D, E and F be the digits
Now, C =  A, B = 2A
F = A + B + C = A + 2A + A = 4A, E = A + B = A + 2A = 3A
and D =  E + F = 3A + 4A = 7A
Since A an d D both are digit, the maximum possible value of A = 1. Therefore, the maximum value of D = 7

Both the given sequences are A.P.
The common difference for the first sequence, d₁ = 4
The common difference for the second sequence, d₂ = 5
The first common term in the two given A.P. is 19.
The common terms will also be in arithmetic progression with common difference
LCM (d₁, d₂) = LCM (4, 5) = 20
Let there be ‘n’ terms in this sequence, then the last term would be ≤ 415
Hence for A.P. of common terms
a + (n – 1) d ≤ 415
⇒ 19 + (n – 1) × 20 ≤ 415
⇒ (n – 1) × 20 ≤ 396 
⇒ (n – 1) = [396 / 20] where [396 / 20] is the greatest integer
⇒ (n –1) = 19
∴ n = 20

Let score of A, B, C and D are a, b, c and d respectively.
Score of A = 72,
∴ a = 72
Also a = 0 .9 × b ⇒ b = a/0.9 = 72 / 0.9 = 80
and b = 1 .25 × c ⇒ c = b / 1.25 = 80 / 1.25 = 64
and e = 0.8 × d ⇒ d = c / 0.8 = 64 / 0.8 = 80
Hence score of D = 80

(√2)¹⁹ 3⁴ 4² 9^m 8ⁿ = 3ⁿ 16^m (∜64)
⇒ 2^19/2 × 3⁴ × 2⁴ × 3²m × 2³n = 3n × 2⁴m × 2^3/2
⇒ 2^{(19/2 + 4 + 3}n) × 3^{(4 + 2}m) = 
On comparing the powers of same bases from both side, we get
19 / 2 + 4 + 3n = 4m + 3 / 2
⇒ 3n = 4m + - 4
⇒ 3n = 4m + 
⇒ n = 
⇒ n =   ...(i)
4 + 2m = n   ...(ii)
From equation (i) and (ii), we get

⇒ 2m = –24
∴ m = –12

Let the two numbers be x and y
∴ x × y = 616
and 
Now, let x³ – y³ = 157k and (x – y)³ = 3k
Since, (x – y)³ = x³ – y³ – 3xy(x – y)
∴ 3k = 157k – 3 × 616(3k)^1/3
⇒ 154k = 3 × 616 × (3k)^1/3 ⇒ k =
⇒ k = 12 × (3k)^1/3 ⇒ k³ = 12³ × 3 × k
⇒ k(k² – 12³ × 3) = 0
∵ k ≠ 0
∴ k² = 3 × 12³
∴ k = 72
⇒ x – y = (3k)^1/3 = (3 × 72)^1/3 = 6
∵ (x + y)² = (x – y)² + 4xy
∴ (x + y)² = 62 + 4 × 616 = 2500
⇒ x + y = 50
Hence, sum of the two numbers = 50

Given, (5.55)x = 1000 ⇒ (5.55)x = 10³
Taking log both the sides, we get
x log10(5.55) = 3 ⇒ log10(5.55) = 3 / x
⇒ log10(10 × 0.555) = 3 / x
⇒ log10(0.555) + 1 = 3 / x  ...(i)
Also given, (0.555)y = 1000
Taking log both the sides, we get
y log10(0.555) = 3 ⇒ log10(0.555) = 3 / y  ...(ii)
From (i) and (ii), we get

0.25 < 2^x < 200
Possible values of x satisfying the above inequality are –2, –1, 0, 1, 2, 3, 4, 5, 6, 7. When x = 0, 1, 2, 4 and 6, 2^x + 2 is divisible by 3 or 4.
Hence, required number of values of x is 5.

Let the other two numbers be y and z.
73yz – 37yz = 720
yz = 20
Minimum possible sum of the squares of the other two numbers would occur when y = z
If y = z, then y² = 20 = z²
Hence, y² + z² = 40

Let the number of girls be 2x and number of boys be x.
Girls getting admission = 0.6 x
Boys getting admission = 0.45 x
Number of students not getting admission = 3x – 0.6x – 0.45x = 1.95x

Alternatively,
Let the no. of boys appearing for the admission test be b % of candidates who get admission.

Here, 120 ≤ n ≤ 240.
120 = 2³ (3)(5) and 240 = 2⁴ (3)(5)
So, the prime factors involved in 120 and 240 are the same.
So, number of co-primes of 240 lying between 120 and 240 = φ(240) – φ(120).

Remainder [N/625]

Let the 4-digit sequence be abcd.
In base 6, this represents 216a + 36b + 6c + d and each of a, b, c, d is less than 6.
In base 10, it represents 1000a + 100b + 10c + d.
Given 4(216a + 36b + 6c + d)
= 1000a + 100b + 10c + d
⇒ 136a = 44b + 14c + 3d ...(A)
By trial a = 1, b = 2, c = 3, d = 2
If a = 2, the LHS = 272
[If we consider b = 5, we need 272 – 220 or 52 from 14c + 3d (c, d)=(2, 8) but 8 is not a proper digit in base 6.
If a= 3, the LHS = 408, while 44b + 14c + 3d can at the most be (44 + 14 + 3) 5 or 305.
∴ There are no other possible values that satisfy (A)]
∴ abcd = 1232 and a + b + c + d = 8

Here, 7⁴ = 2401
∴ 7⁷⁰⁰ = (7⁴)¹⁷⁵ = (2401)¹⁷⁵
Any power of 2401 will end with 1 as the units digit and 0 as the tens digit.
∴ When it is divided by 100, the remainder is 1.

1000 = 2³ × 5³ and 2000 = 2⁴ × 5³
Since LCM (c, a) and LCM (b, c) is 2⁴ × 5³ and LCM (a, b) = 2³ × 5³, so the factor 2⁴ must be present in c.
Hence c = 2⁴ × 5^x, where x ranges from 0 to 3 Therefore, there are four possible values of C.
Since, HCF of (a, b) = K × 5³, it means
a = 2^y × 5³
b = 2^z × 5³
x = 0 to 3, y = 0, then z = 3 → 4 cases.
x = 0 to 3, y = 1, then z = 3 → 4 cases.
x = 0 to 3, y = 2, then z = 3 → 4 cases.
x = 0 to 3, y = 3, then z = 3 → 4 cases.
x = 0 to 3, y = 3, then z = 2 → 4 cases.
x = 0 to 3, y = 3, then z = 1 → 4 cases.
x = 0 to 3, y = 3, then z = 0 → 4 cases.
Hence, total cases = 28.

x = 106⁹⁰ – 49⁹⁰
∵ (xⁿ – aⁿ) is divisible by both (x – a) and (x + a) whenever n is even
⇒ (106⁹⁰ – 49⁹⁰) is divisible by both 57 and 155
57 = 19 × 3
155 = 31 × 5
Therefore, (106⁹⁰ – 49⁹⁰) will be divisible by (19 × 31) = 589 as well.
Also, note that (106⁹⁰– 49⁹⁰) will be odd and options (b) and (c) are even. Hence, they can be rejected.

Let n be xyz and since n is odd z can take only odd values i.e. 1, 3, 5 and 9 Now, x ≤ y and x ≥ z

∴ Total number of elemenls in P = 95.

Solve by option.
Option (a): If the product of the digits is 6. then the factors of 6 are 1,2, 3 and 6. This combination of digits is not suitable. So it is not the answer.
Option (b): If the product of the digits is 8, then the factors of 8 are 1, 2, 4 and 8. So only possible combination is 1, 1, 2, 4.
Hence, the number is 4112. It is suitable for answer.
Similarly, we can check options (c) and (d).

P = 15¹⁰⁰ (1 × 2 × 3 × .... × 100)
= 15¹⁰⁰ × 100!
Highest power of 2 in P = 97 (2 will be deciding factor for number of zeroes because number of lives will be greater than number of zeroes in this number) Q = 25^20×50 (1 × 2 × 3x .... × 50) = 5²⁰⁰⁰ × 50!
Highest power of 2 in θ = 47
So Highest power of 2 in
P²/Q × 10¹⁷⁶⁷ = 2 × 97 + 1767 – 47 = 1914
Hence, number of zeroes = 1914.

x = L.C.M. of (7, 8, 9) – 3 = 504 – 3 = 501
x³ + 2x² + x – 3 = (x – 1) (x + 1) (x + 2) –1
= 500 × 502 × 503 – 1
Remainder when 500 × 502 × 503 – 1 is divided by:
11 = 4
3 = 0
4 = 3
Required remainder = least possible number which when divided by 11, 3 and 4 leavs remainder 4, 0 and 3 respectively.
Such least no. is 15.

In order to maximize the power of 2 in the product, one of the ten numbers has to be 64 as this is the highest two–digit number of the form 2^k, where k is a natural number.
There has to be maximum number of multiples of 8 among the ten numbers. In a set of ten consecutive natural numbers, there can be a maximum of two numbers that will be a multiple of 8.
The possible sets of ten consecutive natural numbers that satisfy the aforementioned conditions are 55 to 64, 56 to 65, 63 to 72 and 64 to 73. The highest power of 2 in the product of any of these sets of ten numbers will be 13.

Here, 343^b = 676
⇒ 7^3b = 26²
Now,7^a = 26
⇒ 7^3b = (7^a)²
⇒ 2a = 3b

Here, we take the 1st option, a delete all perfect squares and perfect cubes, then a total of 22 perfect square will be deleted (1², 2² ........ 22²) and a total of 7 perfect cubes will be deleted (1³ 2³,.... 7³) and Two numbers are common in between them viz. 1⁶ and 2⁶ which are perfect squares as well as perfect cubes Thus, 500 is the (500 – 22 – 7 + 2) = 473 rd term.
So, 476th term = 500 + 3 = 503.

Let the batsman scored a 2's, b 4's and c 6's.
⇒ 2a + 4b + 6c = 100
⇒ a + 2b + 3c = 50. ...(i)
When c = 1, (i) becomes a + 2b = 47
⇒ a = 47 – 2b ....(ii)
Since a ≥ 1 and b ≥ 1, the number of solutions of (ii) is 23.
When c = 2, (i) becomes a + 2b = 44
⇒ a = 44 – 2b ...(iii)
Since a ≥ 1 and b ≥ 1, the number of solutions of (iii) is 21.
When c = 3, (i) becomes a + 2b = 41
⇒ a = 41 – 2b ...(iv)
Since a ≥ 1 and b ≥ 1, the number of solutions of (iv) is 20.
When c = 4, (i) becomes a + 2b = 38
⇒  a = 38 –2b ...(v)
Since a ≥ 1 and b ≥ 1, the number of solutions of (v) is 18.
Thus, we see a pattern emerging.
∴ The total number of ways
= 23 + 21 + 20 + 18 + ... + 3 + 2 = 184.

Put n = 5, in the given relation then, P₅ = 267 Again,
P₆ = P₅ – P₄ + P₃ – P₂
⇒ P₆ = – P₁
P₇ = – P₂,
Similarly, P₈ = – P₃
P₉ = – P₄
P₁₀ = – P₅
The sequence repeats its terms after every 10 terms.
Here, we observe following pattern
P₅₃₁ = P_{(530 + 1)} = P₁ = 211
P₇₅₃ = P_{(750 + 3)} = P₃ = 420
P₉₇₅ = P_{(970 + 5)} = P₅ = 267
So, P₅₃₁ + P₇₅₃ + P₉₇₅ = 211 + 420 + 267 = 898.

Let the two numbers be x and y according to question,
x + y = 600 and x/y = 7/8
∵ x + y = 600 ...(i)
x/y = 7/8 ⇒ 8x - 7y = 0 .....(ii) 
From equation (i) and (ii)
x = 280 and y = 320
∴ LCM of 280 and 320 = 2240

For d = 1, Total = 46
(1, 2, 3, 4, 5), (2, 3, 4, 5, 6) ............(46, 47, 48, 49, 50)
For d = 2, total = 42
(1, 3, 5, 7, 9), (2, 4,6, 8, 10)......... (42, 44, 46, 48, 50)
For d = 3, total = 38
(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) ........ (38, 41, 44, 47, 50)
For d = 12, total = 2
(1, 13, 25, 37, 49) (2, 14, 26, 38, 50)
So total = 46 + 42 + 38 ..........2
Possible APs = 12/2 (2 + 46) = 288.

∵ f(2n) = 1⁴ + 2⁴ + 3⁴ + 4⁴ + 5⁴ +.... + (2n)⁴
⇒ f(2n) = (1⁴ + 3⁴ + 5⁴ + .... + (2n – 1)⁴) + (2⁴ + 4⁴ + 6⁴ + .... + (2n)⁴)
∴ 1⁴ + 3⁴ + 5⁴ + ..... + (2n – 1)⁴
= f(2n) – (2⁴ + 4⁴ + 6⁴ + .... + (2n)⁴)
= f(2n) – 2⁴ × (1⁴ + 2⁴ + 3⁴ + .... + n⁴)
= f(2n) – 16 × f(n)

1080 = 2³ × 3³ × 5¹
(where N = a^pb^qc^r
∴ No. of factors (p + 1) (q + 1) (r + 1)
∴ No. of factors of 1080 (3 + 1) (3 + 1) (1 + 1) = 4 × 4 × 2 = 32
1800 = 2³ × 3² × 5²
(where N =  a^pb^qc^r
∴ No. of factor (p + 1) (q + 1) (r + 1)
∴ Number of factor of 1800 = (3 + 1)(2 + 1)(2 +1)
= 4 × 3 × 3 = 36
∴  HCF of 1080 and 1800 = 2³ × 3² × 5
where N = a^p, b^q, c^r
No. of factors HCl = (p + 1)(q + 1)(r +1)
∴ No. of factors HCF of two numbers = (3 + 1) (2 + 1) (1 + 1)
= 4 × 3 × 2 = 24
So, the required number of divisors
= (32 + 36) – 2 × 24 = 20