Algebra - Number Theory - Previous Year CAT/MBA Questions
You can practice all previous year OMET questions from the topic Algebra - Number Theory. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.
What is the remainder if 19^{20} – 20^{19} is divided by 7?
- A.
5
- B.
1
- C.
6
- D.
0
- E.
3
Answer: Option A
Explanation :
$R\left[\frac{{19}^{20}}{7}\right]$ = $R\left[\frac{{5}^{20}}{7}\right]$ = $R\left[\frac{{25}^{10}}{7}\right]$ = $R\left[\frac{{4}^{10}}{7}\right]$ = $R\left[\frac{{16}^{5}}{7}\right]$ = $R\left[\frac{{2}^{5}}{7}\right]$ = $R\left[\frac{32}{7}\right]$ = 4
$R\left[\frac{{20}^{19}}{7}\right]$ = $R\left[\frac{{(-1)}^{19}}{7}\right]$ = -1
$R\left[\frac{{19}^{20}-{20}^{19}}{7}\right]$ = 4 - (-1) = 5
Hence, option (a).
Workspace:
When expressed in a decimal form, which of the following numbers will be nonterminating as well as non-repeating?
- A.
$\left(\frac{\pi}{2}\right)\left[\frac{1}{\pi}+1\right]-\frac{\pi}{2}$
- B.
sin^{2}1° + sin^{2}2° + … + sin^{2}89°
- C.
$\frac{\sqrt[3]{729}}{3}+\frac{22}{7}$
- D.
$(4-\pi )\left[1+\frac{\pi}{4}+{\left(\frac{\pi}{4}\right)}^{2}+{\left(\frac{\pi}{4}\right)}^{3}+...infiniteterms\right]$
- E.
$\sqrt{2}\left(3\sqrt{2}-\frac{4}{\sqrt{2}}\right)+\sqrt{3}$
Answer: Option E
Explanation :
We need to check options.
Option (a): $\frac{\mathrm{\pi}}{2}\left[\frac{1}{\mathrm{\pi}}+1\right]-\frac{\mathrm{\pi}}{2}$
= $\frac{1}{2}+\frac{\mathrm{\pi}}{2}-\frac{\mathrm{\pi}}{2}$ = $\frac{1}{2}$ = 0.5 (terminating)
Option (b): sin21° + sin22° + … + sin288° + sin289°
= sin21° + sin22° + … + sin245° + … + cos22° + cos21°
= 44 + ½ = 44.5 (terminating)
Option (c): $\frac{\sqrt[3]{729}}{3}+\frac{22}{7}$
= $\frac{9}{3}+\frac{22}{7}$ = 3 + 22/7 (terminating)
Option (d): $\left(4-\mathrm{\pi}\right)[1+\frac{\mathrm{\pi}}{4}+{\left(\frac{\mathrm{\pi}}{4}\right)}^{2}+{\left(\frac{\mathrm{\pi}}{4}\right)}^{3}+...inifiniteterms]$
= $\left(4-\mathrm{\pi}\right)\left[\frac{1}{1-{\displaystyle \frac{\mathrm{\pi}}{4}}}\right]$ = 4 (non-terminating)
Option (e): $\sqrt{2}\left(3\sqrt{2}-\frac{4}{\sqrt{2}}\right)+\sqrt{3}$
= 6 - 4 + √3 = 2 + √3 (non-terminating and non-repeating)
Hence, option (e).
Workspace:
Consider the four variables A, B, C and D and a function Z of these variables, Z = 15A^{2} − 3B^{4} + C + 0.5D. It is given that A, B, C and D must be non-negative integers and that all of the following relationships must hold:
i) 2A + B ≤ 2
ii) 4A + 2B + C ≤ 12
iii) 3A + 4B + D ≤ 15
If Z needs to be maximised, then what value must D take?
- A.
15
- B.
12
- C.
0
- D.
10
- E.
5
Answer: Option B
Explanation :
To maximise Z, we need to maximise A, C and D and minimise B.
Given, 2A + B ≤ 2
A can be either 1 or 0.
Case 1: If A = 1, minimum value of B = 0
From (ii): Maximum value of C = 8
From (iii): Maximum value of D = 12
⇒ Z = 15A^{2} − 3B^{4} + C + 0.5D = 15 – 0 + 8 + 6 = 29
Case 2: If A = 0, minimum value of B = 0
From (ii): Maximum value of C = 12
From (iii): Maximum value of D = 15
⇒ Z = 15A^{2} − 3B^{4} + C + 0.5D = 0 – 0 + 12 + 7.5 = 19.5
We can see that maximum value of Z = 29, when D = 12.
Hence, option (b).
Workspace:
When opening his fruit shop for the day a shopkeeper found that his stock of apples could be perfectly arranged in a complete triangular array: that is, every row with one apple more than the row immediately above, going all the way up ending with a single apple at the top.
During any sales transaction, apples are always picked from the uppermost row, and going below only when that row is exhausted. When one customer walked in the middle of the day she found an incomplete array in display having 126 apples totally. How many rows of apples (complete and incomplete) were seen by this customer? (Assume that the initial stock did not exceed 150 apples.)
- A.
14
- B.
13
- C.
15
- D.
11
- E.
12
Answer: Option E
Explanation :
In the given triangular array, 1st row will have 1 apple, 2nd row will have 2 apples and so on.
∴ n^{th} row will have n apples.
⇒ Total number of apples initially = 1 + 2 + 3 + … + n = $\frac{n(n+1)}{2}$
Initially total number of apples should be greater than 126 but less than or equal to 150
∴ 126 < $\frac{n(n+1)}{2}$ ≤ 150
Only integral value of n satisfying this is 16.
⇒ $\frac{16\times 17}{2}$ = 136
∴ Initially there are 16 rows and 136 apples arranged in a triangular array in these 16 rows.
As the customer observes 126 apples, it would mean that 10 apples were removed.
Now, apples are removed starting from Row 1, which has 1 apple. So apples must have been removed in such a way with 1 apple from Row 1, 2 apples from Row 2, 3 apples from Row 3 and 4 apples from Row 4.
∴ Total 10 apples were removed the first 4 rows.
As out of 16 rows, apples are removed till Row 4, it would mean that only 16 - 4 or 12 rows of apples are visible to the customer.
Hence, option (e).
Workspace:
If x^{2} + x + 1 = 0, then x^{2018 }+ x^{2019} equals which of the following:
- A.
-x
- B.
x
- C.
None of the others
- D.
x - 1
- E.
x + 1
Answer: Option A
Explanation :
x^{2} + x + 1 = 0 …(1)
⇒ x^{2} = - x – 1 …(2)
Multiplying x in (2)
⇒ x^{3} = - x^{2} – x
⇒ x^{3} = 1 …(3) [From (1)]
Now,
x^{2018} + x^{2019} = x^{2018}(1 + x)
= x^{2018} × -x^{2}
= -x^{2020}
= - (x^{3})^{673} × x
= - 1 × x
= - x
∴ x^{2018} + x^{2019 }= - x
Hence, option (a).
Workspace:
We have two unknown positive integers m and n, whose product is less than 100.
There are two additional statement of facts available:
- mn is divisible by six consecutive integers { j, j + 1,...,j + 5 }
- m + n is a perfect square.
Which of the two statements above, alone or in combination shall be sufficient to determine the numbers m and n?
- A.
Statements 1 and 2 together are not suﬃcient, and additional data is needed to answer the question.
- B.
Both statements taken together are suﬃcient to answer the question, but neither statement alone is suﬃcient.
- C.
Each statement alone is suﬃcient to answer the question.
- D.
Statement 2 alone is suﬃcient, but statement 1 alone is not suﬃcient to answer the question.
- E.
Statement 1 alone is suﬃcient, but statement 2 alone is not suﬃcient to answer the question.
Answer: Option B
Explanation :
Looking at the initial statement , we know that mn < 100.
Looking at statement I ,we get to know that the product mn has to be a multiple of 10 since it is divisible by 6 consecutive integers .So the product can be either 10, 20, 30 ….90
Now of all these numbers only 60 is divisible by 6 consecutive numbers i.e. numbers 1 to 6.60 can be expressed as a product of 2 nos. in the following ways : 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12, 6 ×10
So from statement I alone we cannot determine values of m and n. Looking at statement II alone determine values of m and n as the only information provided to us is that “m + n” is a perfect square. So we can have numerous possibilities for m and n [e.g (7, 9), (2, 7), (1, 3) etc ]
Combining both statements out of (1, 60), (2, 30), (3, 20), (4,15), (5, 12), (6, 10), the only pair of values such that “m+n” is a perfect square is (6, 10). Hence both statements are required to answer the question.
Hence, option (b).
Workspace:
The number 37^{371} – 26^{371} is divisible by:
- A.
10
- B.
11
- C.
12
- D.
15
Answer: Option B
Explanation :
37^{371} − 26^{371} is of the form a^{n} − b^{n}
and, a^{n} − b^{n} is always divisible by a − b
Thus, the expression will be divisible by 37 − 26 = 11
Hence, option (b).
Workspace:
Find the value of the expression: 10 + 10^{3} + 10^{6} + 10^{9}
- A.
1010101010
- B.
1001000010
- C.
1001000110
- D.
1001001010
- E.
100010001010
Answer: Option D
Explanation :
10 + 10^{3} + 10^{6} + 10^{9} =
10 +
1000 +
1000000 +
1000000000 +
= 1001001010
Hence, option (d).
Workspace:
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 23^{2} is 529).
- A.
9
- B.
10
- C.
11
- D.
12
- E.
None of the above
Answer: Option B
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 X^{th} day of a month. For example, he spends exactly £5 on the 5^{th} 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 2^{N} 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
Explanation :
Sum of any four consecutive numbers = x + x + 1 + x + 2 + x + 3 = 4x + 3.
This sum is never divisible by 4. But 2^{N} 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 2^{N} = 64 = 2^{6}.
∴ 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
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 = (11^{p+7})(7^{q-2})(5^{r+1})(3^{s}) 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
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 3^{s} to be a perfect cube, minimum value of s = 3.
For 5^{r+1} to be a perfect cube, minimum value of r = 2.
For 7^{q-2} to be a perfect cube, minimum value of q = 2.
For 11^{p+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
Explanation :
500 = 2^{2} × 5^{3}
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 10^{10}.
∴ 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 $\frac{{a}^{3}+{b}^{3}+{c}^{3}+3abc}{{(a+b+c)}^{2}}$?
- A.
0
- B.
1
- C.
2
- D.
3
- E.
4
Answer: Option C
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,
= $\frac{{(b-1)}^{3}+{b}^{3}+{(b+1)}^{3}+3(b-1)b(b+1)}{{(b-1+b+b+1)}^{2}}$
= $\frac{6{b}^{3}+3b}{{\left(3b\right)}^{2}}$
= $\frac{2{b}^{2}+1}{3b}$
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
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 3.
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
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
$\therefore (x-1)=\frac{abc...999000}{N!}$
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) = M^{2} – NM can take four of the five given values.
Hence, there are five possible equations – one for each option.
Consider option 1: M^{2} – 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
M^{2} – 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
M^{2} – 17M – 200 = 0 i.e. M = 25 or M = −8. Here, M = 25 is valid.
Option 4:
For N = 16, the equation becomes
M^{2} – 16M – 225 = 0 i.e. M = 25 or M = −9. Here, M = 25 is valid.
Option 5:
For N = 17, the equation becomes
M^{2} – 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 M^{2} – NM – 180 = 0
Hence, M(M – N) can never be 180.
Hence, option 2.
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
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 2.
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
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 3.
Workspace:
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