Algebra - Number Theory - Previous Year IPM/BBA Questions
The best way to prepare for Algebra - Number Theory is by going through the previous year Algebra - Number Theory questions for IPMAT - Indore. Here we bring you all previous year Algebra - Number Theory IPMAT - Indore questions along with detailed solutions.
Click here for previous year questions of other topics.
It would be best if you clear your concepts before you practice previous year Algebra - Number Theory questions for IPMAT - Indore.
ipm
The remainder when 1! + 2! + 3! +.... + 95! is divided by 15 is.
Answer: 3
Text Explanation :
Here, factorials of number 5 and above are always divisible by 15, hence we can ignore them.
So the question simplifies to finding remainder of 1! + 2! + 3! + 4! when divided by 15.
= Remainder of (1 + 2 + 6 + 24) / 15
= Remainder of (33) / 15
= 3
Hence, 3.
Workspace:
The polynomial 4x10 - x9 + 3x8 – 5x7 + cx6 + 2x5 – x4 + x3 – 4x2 + 6x – 2 when divided by x – 1 leaves a remainder 2. Then the value of c + 6 is
Answer: 5
Text Explanation :
Concept: Remainder of f(x) when divided by x - a is f(a).
Here f(x) = 4x10 - x9 + 3x8 – 5x7 + cx6 + 2x5 – x4 + x3 – 4x2 + 6x – 2
It has to be divided by x - 1.
∴ Remainder will be f(1)
⇒ f(1) = 4 × 110 - 19 + 3 × 18 – 5 × 17 + c × 16 + 2 × 15 – 14 + 13 – 4 × 12 + 6 × 1 – 2 = 2
⇒ 4 - 1 + 3 – 5 + c + 2 – 1 + 1 – 4 + 6 – 2 = 2
⇒ c + 3 = 2
⇒ c = - 1
∴ c + 6 = -1 + 6 = 5
Hence, 5.
Workspace:
The total number of positive integer solutions of 21 < a + b + c < 25 is
Answer: 1160
Text Explanation :
We know number of ways of distributing n identical objects among r different groups such that each groups gets at least one object is n-1Cr-1.
Now,
Case 1: a + b + c = 21.
∴ We need to distribute 21 among 3 different variables a, b & c.
Number of ways of doing this = 21-1C3-1 = 20C2 = 190
Case 2: a + b + c = 22.
Number of ways of doing this = 22-1C3-1 = 21C2 = 210
Case 3: a + b + c = 23.
Number of ways of doing this = 23-1C3-1 = 22C2 = 231
Case 4: a + b + c = 24.
Number of ways of doing this = 24-1C3-1 = 23C2 = 253
Case 5: a + b + c = 25.
Number of ways of doing this = 25-1C3-1 = 24C2 = 276
∴ Total number of ways = 190 + 210 + 231 + 253 + 276 = 1160
Hence, 1160.
Workspace:
In a chess tournament there are 5 contestants. Each player plays against all the other exactly once. No game results in a draw. The winner in a game gets one point and the loser gets zero point. Which of the following sequences cannot represent the scores of the five players?
- (a)
2, 2, 2, 2, 2
- (b)
3, 3, 2, 1, 1
- (c)
3, 2, 2, 2, 1
- (d)
4, 4, 1, 1, 0
Answer: Option D
Text Explanation :
Workspace:
Let p be a positive integer such that the unit digit of p3 is 4. What are the possible unit digits of (p+3)3?
- (a)
1, 3
- (b)
1, 3, 7
- (c)
4, 7
- (d)
1, 7, 9
Answer: Option A
Text Explanation :
Workspace:
Let [x] denote the greatest integer not exceeding x and {x} = x – [x]. If is a natural number, then the sum of all values of x satisfying the equation 2[x] = x + n{x} is
- (a)
n(n + 2)/2
- (b)
3/2
- (c)
n
- (d)
n(n + 1)/2
Answer: Option A
Text Explanation :
Workspace:
If the five-digit number abcde is divisible by 6 , then which of the following numbers is not necessarily divisible by 6?
- (a)
edcba
- (b)
eee
- (c)
bbadcacede
- (d)
cdbae
Answer: Option A
Text Explanation :
For a number to be divisible by 6, the sum of its digits should be divisible by 3 and it's units digit should be even.
Since, abcde is divisible by 6,
a + b + c + d + e = multiple of 3, and
e is an even number
Option (a): edcba
Here the units digit is a which is not necessarily even, hence edcba is not definitely divisible by 6.
Option (b): eee
Here units digit e is even and sum of the digits = 3e which is a multiple of 3, hence eee is definitely divisible by 6.
Option (c): bbadcacede
Here units digit e is even and sum of the digits = 2(a + b + c + d + e) which is a multiple of 3, hence eee is definitely divisible by 6.
Option (d): cdbae
Here units digit e is even and sum of the digits = (a + b + c + d + e) which is a multiple of 3, hence eee is definitely divisible by 6.
Hence, option (a).
Workspace:
When the square of the difference of two natural numbers is subtracted from the square of the sum of the same two numbers and the result is divided by four, we get
- (a)
the product of the LCM and HCF of the two numbers
- (b)
the HCF of the two numbers
- (c)
the LCM of the two numbers
- (d)
the square of the product of the two numbers
Answer: Option A
Text Explanation :
Let the two numbers be hx and hy, where h is the HCF of thest two numbers and LCM of these numbers = hxy.
Now, (hx + hy)2 - (hx - hy)2
= (hx)2 + (hy)2 + 2h2xy - ((hx)2 + (hy)2 - 2h2xy)
= 4h2xy
= 4 × h × hxy
= 4 × HCF × LCM
Hence, option (a).
Workspace:
The number of positive integers that divide (1890) × (130) × (170) and are not divisible by 45 is:
Answer: 320
Text Explanation :
We first need to find all factors of (1890) × (130) × (170) and then remove those factors which are divisible by 45.
(1890) × (130) × (170) = (2 × 33 × 5 × 7) × (13 × 2 × 5) × (17 × 2 × 5)
⇒ 1890 × 130 × 170 = 23 × 33 × 53 × 7 × 13 × 17
∴ Total factors = (3 + 1)(3 + 1)(3 + 1)(1 + 1)(1 + 1)(1 + 1) = 512
Now for factors that are divisible by 45, least power of 3 should be 2 and that of 5 should be 1. Powers of 2, 7, 13 and 17 can be anything.
∴ Total factors which are divisible by 45 = 4 × 2 × 3 × 2 × 2 × 2 = 192
⇒ Number of factors which are not divisible by 45 = 512 - 192 = 320.
Hence, 320.
Workspace:
What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? (Weights can be placed only on the left pan)
Answer: 7
Text Explanation :
If weight can be placed only on one pan, we need to take weights which are powers of 2.
Weights required are 1 kg, 2 kg, 4 kg, 8 kg, 16 kg, 32 kg and 64 kg i.e., 7 weights.
Hence, 7.
Workspace:
The unit digit in (743)85 – (525)37 + (987)96 is ________
- (a)
9
- (b)
3
- (c)
1
- (d)
5
Answer: Option A
Text Explanation :
Unit's digit of (743)85 = Unit's digit of 385
= Unit's digit of (34)21 × 3 = 1 × 3 = 3 [Unit's digit of 34 = 1]
Units's digit of (525)37 = Units's digit of 537 = 5
Units's digit of (987)96 = Units's digit of 796
= Units's digit of (74)24 = 1 [Unit's digit of 74 = 1]
∴ Unit's digit of (743)85 – (525)37 + (987)96 = 3 - 5 + 1 = -1 = 9
Hence, option (a).
Workspace:
The highest possible value of the ratio of a four-digit number and the sum of its four digits is
- (a)
1000
- (b)
277.75
- (c)
900.1
- (d)
999
Answer: Option A
Text Explanation :
The highest possible value of the ratio of a four-digit number and the sum of its four digits is
Let the number be 'abcd' = 1000a + 100b + 10c + d
We have to find the highest possible value of
(1000a + 100b + 10c + d) : (a + b + c + d)
This is possible when a is highest and b, c and d are lowest
i.e., a = 9 and b = c = d = 0
∴ Highest possible ratio = 9000 : 9 = 1000
Hence, option (a).
Workspace:
The number of 5-digit numbers consisting of distinct digits that can be formed such that only odd digits occur at odd places is
- (a)
5250
- (b)
6240
- (c)
2520
- (d)
3360
Answer: Option C
Text Explanation :
Let the five digit number be 'abcde'
All digits are distinct and digits a, c and e must be odd
There are 5 odd digits i.e., 1, 3, 5, 7 and 9.
a can take any of these 5 digits in 5 ways.
c can take any of the remaining 4 digits in 4 ways.
e can take any of the remaining 3 digits in 3 ways.
∴ Number of ways of choosing values for a, c and e = 5 × 4 × 3 = 60 ways.
b and d can take distinct values out of 0, 2, 4, 6, 8 and remaining 2 odd digits in 7 × 6 = 42 ways.
∴ Total number of ways of forming the requried 5 digit number = 60 × 42 = 2520 ways.
Hence, option (c).
Workspace:
In a division problem, product of quotient and the remainder is 24 while their sum is 10. If the divisor is 5 then dividend is __________.
Answer: 34
Text Explanation :
Let the dividend be N, Quotient be Q and Remainder be R.
Given, Divisor = 5
∴ N = 5 × Q + R
Given, QR = 24 and Q + R = 10
Since Q and R are integers, the only possible value for Q and R will be 6 or 4 in any order.
Here, the divisor is 5 and hence remainder cannot be greater than 5, hence R = 4 and Q = 6.
⇒ N = 5 × 6 + 4 = 34.
Hence, 34.
Workspace:
In a class, students are assigned roll numbers from 1 to 140. All students with even roll numbers opted for cricket, all those whose roll numbers are divisible by 5 opted for football, and all those whose roll numbers are divisible by 3 opted for basketball. 'The number of students who did not opt for any of the three sports is
- (a)
102
- (b)
38
- (c)
98
- (d)
42
Answer: Option B
Text Explanation :
Number of students who opted for cricket = Number of number which are divisible by 2 = Quotient of 140/2 = 70.
Number of students who opted for football = Number of number which are divisible by 5 = Quotient of 140/5 = 28.
Number of students who opted for basketball = Number of number which are divisible by 3 = Quotient of 140/3 = 46.
Number of students who opted for cricket and football = Number of number which are divisible by 2 and 5 = Quotient of 140/10 = 14.
Number of students who opted for basketball and football = Number of number which are divisible by 3 and 5 = Quotient of 140/15 = 9.
Number of students who opted for cricket and basketball = Number of number which are divisible by 2 and 3 = Quotient of 140/6 = 23.
Number of students who opted for cricket and football and basketball = Number of number which are divisible by 2, 5 and 3 = Quotient of 140/30 = 4.
∴ C ∪ F ∪ B = C + F + B - C ∩ F - F ∩ B - B ∩ C + C ∩ F ∩ B
⇒ C ∪ F ∪ B = 70 + 28 + 46 - (14 + 9 + 23) + 4
⇒ C ∪ F ∪ B = 144 - 46 + 4 = 102
∴ Number of students who opted for none of the three sports = 140 - 102 = 38.
Hence, option (b).
Workspace:
The number of pairs of integers whose sums are equal to their products is
Answer: 2
Text Explanation :
Workspace:
You have been asked to select a positive integer N which is less than 1000 , such that it is either a multiple of 4, or a multiple of 6, or an odd multiple of 9. The number of such numbers is
Answer: 388
Text Explanation :
Workspace:
The maximum value of the natural number n for which 21n divides 50! is
- (a)
6
- (b)
7
- (c)
8
- (d)
9
Answer: Option C
Text Explanation :
21 = 3 × 7
Now, to calculate highest power of a prime number p in N!, we add all the quotients when N is successively divided by p.
So, highest power of 3 in 50! is:
Q(50/3) = 16
Q(16/3) = 5
Q(5/3) = 1
∴ Highest power of 3 in 50! = 16 + 5 + 1 = 22
So, highest power of 7 in 50! is:
Q(50/7) = 7
Q(7/7) = 1
∴ Highest power of 7 in 50! = 7 + 1 = 8
So when 50! is written in prime factorised form it will be:
⇒ 50! = 322 × 78 [There will be power of other prime numbers as well but that is immaterial for this question]
⇒ 50! = 314 × 38 × 78
⇒ 50! = 314 × (3 × 7)8
⇒ 50! = 314 × 218
∴ Highest power of 21 in 50! is 8, hence 218 can divided 50!.
Hence, option (c).
Workspace:
The remainder when is divided by 9 is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Answer: Option B
Text Explanation :
Workspace:
Placing which of the following two digits at the right end of 4530 makes the resultant six digit number divisible by 6,7 and 9?
- (a)
96
- (b)
78
- (c)
42
- (d)
54
Answer: Option A
Text Explanation :
Workspace:
If a, b, c are real numbers a2 + b2 + c2 = 1, then the set of values ab + bc + ca can take is:
- (a)
[-1, 2]
- (b)
[-1/2, 2]
- (c)
[-1, 1]
- (d)
[-1/2, 1]
Answer: Option D
Text Explanation :
Workspace:
There are numbners a1, a2, a3,…,an each of them being +1 or -1. If it is known that a1a2 + a2a3 + a3a4 + …an−1an + ana1 = 0 then
- (a)
n is a multiple of 2 but not a multiple of 4
- (b)
n is a multiple of 3
- (c)
n can be any multiple of 4
- (d)
The only possible value of n is 4
Answer: Option C
Text Explanation :
Workspace: