Algebra - Number Theory - Previous Year CAT/MBA Questions
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X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers (408X)63 and (789Y)85 are the same. What will be the possible value(s) of (X + Y)?
Example: If M = 3 then the digit at unit's place of the number (2M) is 3 (as the number is 23) and the digit at unit's place of the number (2M)2 is 9 (as 232 is 529).
- (a)
9
- (b)
10
- (c)
11
- (d)
12
- (e)
None of the above
Answer: Option B
Text Explanation :
For various powers the units digit of a number always a cycle of 4 terms.
We need unit’s digits of (X)63 and (Y)85 to be same
Unit’s digit of X63 will be same as unit’s digit of X3
Similarly, unit’s digit of Y85 is same as unit’s digit of Y1.
∴ unit’s digits of (X)3 should be same as units digit of Y.
This is possible (from the table) when (X, Y) = (2, 8) or (8, 2) or (7, 3) or (3, 7).
In any of these cases X + Y = 10.
Hence, option (b).
Workspace:
David has an interesting habit of spending money. He spends exactly £X on the Xth day of a month. For example, he spends exactly £5 on the 5th of any month. On a few days in a year, David noticed that his cumulative spending during the last 'four consecutive days' can be expressed as 2N where N is a natural number. What can be the possible value(s) of N?
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
N can have more than one value
Answer: Option B
Text Explanation :
Sum of any four consecutive numbers = x + x + 1 + x + 2 + x + 3 = 4x + 3.
This sum is never divisible by 4. But 2N is always divisible by 4 (for N ≥ 2).
∴ The four consecutive days do not fall in the same month. Let’s look at the various possibilities.
Now, there are 4 possibilities for number of days in a month.
The only possibility for cumulative spending during the last 'four consecutive days' to be expressed as 2N = 64 = 26.
∴ N = 6
Hence, option (b).
Workspace:
An institute has 5 departments and each department has 50 students. If students are picked up randomly from all 5 departments to form a committee, what should be the minimum number of students in the committee so that at least one department should have representation of minimum 5 students?
- (a)
11
- (b)
15
- (c)
21
- (d)
41
- (e)
None of the above
Answer: Option C
Text Explanation :
The five departments of each department has 50 students.
We have to minimise the number of students in the committee so that at least one department should have representation of minimum 5 students
Let us maximise the number of students in the committee so that no department has representation of 5 students. i.e., any department can have maximum 4 students.
∴ If we select 4 students from each department i.e. a total of 20 students, still no department will have representation of 5 students.
Now if we select one more student, we will have representation of 5 students from at least 1 department.
∴ Minimum number of students in the committee so that at least one department should have representation of minimum 5 students is 21.
Hence, option (c).
Workspace:
If N = (11p+7)(7q-2)(5r+1)(3s) is a perfect cube, where p, q, r and s are positive integers, then the smallest value of p + q + r + s is:
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
9
Answer: Option E
Text Explanation :
In order for N to be a perfect cube, every prime number should be a power of multiple of 3 i.e., 0, 3 , 9 …
Also, we need to consider minimum possible values of p, q, r and s. (p, q, r and s > 0)
For 3s to be a perfect cube, minimum value of s = 3.
For 5r+1 to be a perfect cube, minimum value of r = 2.
For 7q-2 to be a perfect cube, minimum value of q = 2.
For 11p+7 to be a perfect cube, minimum value of p = 2.
So, smallest value of p + q + r + s = 2 + 2 + 2 + 3 = 9
Hence, option (e).
Workspace:
For two positive integers a and b, if (a + b)(a+b) is divisible by 500, then the least possible value of a × b is:
- (a)
8
- (b)
9
- (c)
10
- (d)
12
- (e)
None of the above
Answer: Option B
Text Explanation :
500 = 22 × 53
For (a + b)(a+b) to be divisible by 500, it should have power of 2 as well as 5 in it.
∴ a + b should have power of both 2 as well as 5.
The least such number is 1010.
∴ a + b = 10.
∴ a × b is minimum when a = 9 and b = 1 (or vice-versa)
Hence, the least possible value is 9.
Hence, option (b).
Workspace:
If a, b and c are 3 consecutive integers between –10 to +10 (both inclusive), how many integer values are possible for the expression ?
- (a)
0
- (b)
1
- (c)
2
- (d)
3
- (e)
4
Answer: Option C
Text Explanation :
If a, b and c are three consecutive numbers
⇒ a = b – 1 and c = b + 1
Substituting this in the expression we get the expression we get,
=
=
=
Only for b = ±1 is the above expression an integer.
Hence, option (c).
Workspace:
An ascending series of numbers satisfies the following conditions:
- When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
- When divided by 11, the numbers leave no remainder.
The 6th number in this series will be:
- (a)
242
- (b)
2882
- (c)
3542
- (d)
4202
- (e)
None of the above
Answer: Option C
Text Explanation :
LCM (3, 4, 5, 6) = 60
From (i), numbers are of the form 60k + 2.
From (ii), numbers are of the form 11m.
11m = 60k + 2 = 55k + (5k + 2)
11 divides (5k + 2)
Units digit of (5k + 2) is 2 or 7
So, values of (5k + 2) are 11 × 2, 11 × 7, 11 × 12, 11 × 17, 11 × 22, 11 × 27, … and so on.
We need to find the 6th number of the ascending series.
∴ 5k + 2 = 11 × 27 ⇒ k = 59
The required number = 60 × 59 + 2 = 3542
Hence, option (c).
Workspace:
If the last 6 digits of [(M)! – (N)!] are 999000, which of the following option is not possible for (M) × (M – N)?
Both (M) and (N) are positive integers and M > N. (M)! is factorial M.
- (a)
150
- (b)
180
- (c)
200
- (d)
225
- (e)
234
Answer: Option B
Text Explanation :
M and N are positive integers such that M > N
∴ M! – N! = abc…999000
∴ [M(M − 1)(M − 2)……N!] – N! = abc…999000
∴ N!{[M(M − 1) (M − 2)……] − 1} = abc…999000
Let the term in the square bracket be x.
Since M is a positive integer, the term in the square brackets i.e., x is also a positive integer, and hence, (x – 1) is also a positive integer.
∴ N!(x – 1) = abc…999000
Hence, the maximum number of zeroes in N! is 3.
∴ N! ≤ 19 (because from 20! onwards, each factorial has atleast 4 zeroes)
Now, there are 4 possible ranges for N:
- N = 0 to 4 (no zeroes in N!)
- N = 5 to 9 (1 zero in N!)
- N = 10 to 14 (2 zeroes in N!)
- N = 15 to 19 (3 zeroes in N!)
Consider case 3, where there are 2 zeroes in N!
Since N!(x – 1) = abc…999000, the third zero on the LHS should come from (x – 1).
For this, x has to be of the form pqrs…1 i.e., x has to be an odd number.
Now, there are two possibilities:
1. M = N + 1
Here, M! – N! = (M × N!) – N! = N!(M – 1)
In this case, M = x
Since N = 10 to 14 and M – 1 should end in 0, M = 11 and N = 10
Hence, M(M – N) = 11(11 – 10) = 11
This does not tally with any of the options.
Hence, M ≠ N + 1
2. There is at least one integer between M and N
Hence, x comprises a series of consecutive integers (at least two as explained above) multiplied with each other. So, there has to be at least one even number in this series.
Hence, x can never be odd.
Hence, the third zero on the LHS can never come from (x – 1).
Hence, there cannot be 2 zeroes in N!. Similarly, it can be proved that there cannot be 1 zero or no zeroes in N!.
Hence, N! has three zeroes i.e. 15 ≤ N ≤ 19
Now, M(M – N) = M2 – NM can take four of the five given values.
Hence, there are five possible equations – one for each option.
Consider option 1: M2 – NM – 150 = 0
Here, for N = 15 to 19, see if there exists a value of M (> N) that gives positive integral roots in this equation.
For N = 19, the equation becomes
M2 – 19M – 150 = 0 i.e. M = 25 or M = −6. Here, M = 25 is valid.
Similarly, consider each option.
Option 3:
For N = 17, the equation becomes
M2 – 17M – 200 = 0 i.e. M = 25 or M = −8. Here, M = 25 is valid.
Option 4:
For N = 16, the equation becomes
M2 – 16M – 225 = 0 i.e. M = 25 or M = −9. Here, M = 25 is valid.
Option 5:
For N = 17, the equation becomes
M2 – 17M – 234 = 0 i.e. M = 26 or M = −9. Here, M = 26 is valid.
However, no value of N from 15 to 19 gives an integral solution for M in M2 – NM – 180 = 0
Hence, M(M – N) can never be 180.
Hence, option (b).
Note: There is a logical flaw in this question as the difference of two factorials can never end in 999000. The actual question should have had the last six digits as abc000.
Workspace:
A three-digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three-digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions?
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
None of the above
Answer: Option B
Text Explanation :
Units digit of the number must be 0.
Let 100x + 10y is the number such that x > y.
New number obtained by changing the digits is also divisible by 10.
So, only x and y are to be interchanged
∴ New number is of the form = 100y + 10x
Difference = 90x – 90y = 90(x – y)
For the difference to be divisible by 4,
(x – y) has to be divisible by 4.
(x – y) = 4 or 8
So, y = 1 to 5
For y = 1 to 5, x = (1+4) to (5+4) i.e., 5 to 9
One more possibility for y = 1 is x = 9.
Thus, in all 6 numbers satisfy the given conditions.
Hence, option (b).
Workspace:
The business consulting division of TCS has overseas operations in 3 locations: Singapore, New York and London. The Company has 22 analysts covering Singapore, 28 covering New York and 24 covering London. 6 analysts cover Singapore and New York but not London, 4 analysts cover Singapore and London but not New York, and 8 analysts cover New York and London but not Singapore. If TCS has a total of 42 business analysts covering at least one of the three locations: Singapore, New York and London, then the number of analysts covering New york alone is:
- (a)
14
- (b)
28
- (c)
5
- (d)
7
Answer: Option D
Text Explanation :
Using the data we can have following Venn diagram.
From the given data, we can form the following equations:
a + b + c + d = 42 – 18 = 24
⇒ a + b + c + d = 24 … (i)
a + b + c + 2(18) + 3d = 22 + 28 + 24 = 74 ⇒ a + b + c + 3d = 42 … (ii)
Solving (i) and (ii), d = 7
∴ 6 + 7 + 8 + b = 28 ⇒ b = 7
Hence, option (d).
Workspace:
If the product of the integers a, b, c and d is 3094 and if 1 < a < b < c < d, what is the product of b and c?
- (a)
26
- (b)
91
- (c)
133
- (d)
221
Answer: Option B
Text Explanation :
3094 = 2 × 7 × 13 × 17 = a × b × c × d
As 1 < a < b < c < d
a = 2, b = 7, c = 13, d = 17
∴ b × c = 7 × 13 = 91
Hence, option (b).
Workspace:
If the product of n positive integers is nn, then their sum is
- (a)
a negative integer
- (b)
equal to n
- (c)
equal to n +
- (d)
never less than n2
Answer: Option D
Text Explanation :
As all the numbers are positive integers their sum cannot be negative. Thus option 1 is not possible.
Also, as all the numbers are positive integers their sum cannot be a fraction. Thus option 3 is not possible.
Hence, option (d).
Workspace:
Three Vice Presidents (VP) regularly visit the plant on different days. Due to labour unrest, VP (HR) regularly visits the plant after a gap of 2 days. VP (Operations) regularly visits the plant after a gap of 3 days. VP (Sales) regularly visits the plant after a gap of 5 days. The VPs do not deviate from their individual schedules. CEO of the company meets the VPs when all the three VPs come to the plant together. CEO is one leave from January 5th to January 28th, 2012. Last time CEO met the VPs on January 3, 2012. When is the next time CEO will meet all the VPs?
- (a)
February 6, 2012
- (b)
February 7, 2012
- (c)
February 8, 2012
- (d)
February 9, 2012
- (e)
None of the above
Answer: Option C
Text Explanation :
VP (HR) visits on every third day, VP (Operations) visits on every fourth day and VP (Sales) visits on every sixth day.
Hence, all of them will visit together on every twelfth day.
Now, all VPs visited together on January 3, 2012.
Hence, they will visit on, 15th January, 27th January, 8th February and so on.
Hence, option (c).
Workspace:
If k is an integer and 0.0010101 × 10k is greater than 1000, what is the least possible value of k?
- (a)
4
- (b)
5
- (c)
6
- (d)
7
Answer: Option C
Text Explanation :
We need 0.0010101 × 10k > 1000
∴ 1010.1 × 10–6 × 10k > 1000
∴ 1010.1 × 10–6 + k > 1000
∵ 1010.1 > 1000, 10–6 + k = 1
∴ k – 6 = 0
∴ k = 6
Hence, option (c).
Workspace:
A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is:
- (a)
220
- (b)
600
- (c)
240
- (d)
None of the above
Answer: Option D
Text Explanation :
The five digit number will either have 0 or it will not have 0.
Case i:
0 is included.
The five digits must be 0, 1, 2, 4, 5 (as the number has to be divisible by 3)
The count of five-digit numbers = 4 × 4 × 3 × 2 × 1 = 96
Case ii:
0 is excluded.
In this case, the five digits should be 1, 2, 3, 4 and 5.
Number of five digit numbers = 5 × 4 × 3 × 2 × 1 = 120
∴ Required count = 120 + 96 = 216
Hence, option (d).
Workspace:
If 2, a, b, c, d, e, f and 65 form an arithmetic progression, find out the value of ‘e’.
- (a)
48
- (b)
47
- (c)
41
- (d)
None of the above
Answer: Option B
Text Explanation :
65 is the eighth term of the AP.
∴ 2 + (8 – 1)r = 65, where r is the common difference.
∴ r = 9
e is the sixth term.
∴ e = 2 + 5(9) = 47
Hence, option (b).
Workspace:
How many positive integers ‘n’ can we form using the digits 3, 4, 4, 5, 6, 6, 7 if we want ‘n’ to exceed 6,000,000?
- (a)
320
- (b)
360
- (c)
540
- (d)
720
Answer: Option C
Text Explanation :
As n has to exceed 6,000,000, the first digit of n can be 6 or 7.
Case (i):
The first digit of n is 6. Then the other six digits are to be chosen from 3, 4, 4, 5, 6, 7.
They can be arranged in = 360 ways
Case (ii):
The first digit of n is 7.
Then the other six digits are to be chosen from 3, 4, 4, 5, 6, 6.
They can be arranged in = 180 ways
∴ Total number of arrangements = 360 + 180 = 540
Hence, option (c).
Workspace:
The smallest perfect square that is divisible by 7!
- (a)
44100
- (b)
176400
- (c)
705600
- (d)
19600
Answer: Option B
Text Explanation :
7! = 24 × 32 × 5 × 7
The required perfect square should be divisible by 16, 9, 5 and 7.
16 and 9 are already squares. Since 5 and 7 are not perfect squares, we multiply 7! by 5 and 7 to make it a perfect square.
So required perfect square =24 × 32 × 52 × 72 = 176400
Hence, option (b).
Workspace:
In a Green view apartment, the houses of a row are numbered consecutively from 1 to 49. Assuming that there is a value of ‘x’ such that the sum of the numbers of the houses preceding the house numbered ‘x’ is equal to the sum of the numbers of the houses following it. Then what will be the value of ‘x’?
- (a)
21
- (b)
30
- (c)
35
- (d)
42
Answer: Option C
Text Explanation :
Sum of numbers before x = sum of numbers after x
∴ 1 + 2 + 3 + … + (x – 1) = (x + 1) + (x + 2) + … + 49
Adding (1 + 2 + 3 + 4 + … + (x – 1) + x) on both sides
∴ 2(1 + 2 + 3 + … +(x – 1)) + x = (1 + 2 + 3 + … + 49)
∴ + x =
∴ x2 = 49 × 25
∴ x = 7× 5 = 35
Hence, option (c).
Workspace:
How many subsets of {1, 2, 3 … 11} contain at least one even integer?
- (a)
1900
- (b)
1964
- (c)
1984
- (d)
2048
Answer: Option C
Text Explanation :
Total number of subsets = 211
Total number of subsets without any even integer = 26
Hence, required number of subsets
= 211 – 26
= 2048 – 64
= 1984
Hence, option (c).
Workspace:
The number of distinct terms in the expansion of (X + Y + Z + W)30 are:
- (a)
4060
- (b)
5456
- (c)
27405
- (d)
46376
Answer: Option B
Text Explanation :
All the terms in the expansion of (X + Y + Z + W)30 are of the form k.Xa.Yb.Zc.Wd
Where a + b + c + d = 30
∴ The number of terms in the expansion is equal to the number of non-negative integer solutions of the equation a + b + c + d = 30
The number of solutions of this equation is given by
n + r – 1Cr – 1, where n = 30, r = 4
∴ Number of solutions = 33C3 = 5456
Hence, option (b).
Workspace:
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