Algebra - Number Theory - Previous Year CAT/MBA Questions

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).

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): $\left(4-\mathrm{\pi }\right)[1+\frac{\mathrm{\pi }}{4}+{\left(\frac{\mathrm{\pi }}{4}\right)}^2+{\left(\frac{\mathrm{\pi }}{4}\right)}^3+... inifinite terms]$
= $\left(4-\mathrm{\pi }\right)\left[\frac{1}{1-{\displaystyle \frac{\mathrm{\pi }}{4}}}\right]$ = 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).

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 = 15A² − 3B⁴ + 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 = 15A² − 3B⁴ + 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).

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).

Answer: Option A

Explanation :

x² + x + 1 = 0   …(1)

⇒ x² = - x – 1   …(2)

Multiplying x in (2)
⇒ x³ = - x² – x

⇒ x³ = 1   …(3) [From (1)]

Now, 

x²⁰¹⁸ + x²⁰¹⁹ = x²⁰¹⁸(1 + x)

= x²⁰¹⁸ × -x²

= -x²⁰²⁰

= - (x³)⁶⁷³ × x

 = - 1 × x

= - x

∴ x²⁰¹⁸ + x²⁰¹⁹ = - x

Hence, option (a).

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).

Answer: Option B

Explanation :

37³⁷¹ − 26³⁷¹ is of the form aⁿ − bⁿ 

and, aⁿ − bⁿ  is always divisible by a − b

Thus, the expression will be divisible by 37 − 26 = 11

Hence, option (b).

Answer: Option D

Explanation :

10 + 10³ + 10⁶ + 10⁹ = 
                    10 + 
                1000 + 
          1000000 + 
    1000000000 +
=  1001001010

Hence, option (d).

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).

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 = 2⁶.

∴ N = 6

Hence, option (b).

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).

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).

Answer: Option B

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).

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,

= $\frac{{(b-1)}^3+b^3+{(b+1)}^3+3(b-1)b(b+1)}{{(b-1+b+b+1)}^2}$

= $\frac{6b^3+3b}{{\left(3b\right)}^2}$

= $\frac{2b^2+1}{3b}$

Only for b = ±1 is the above expression an integer.

Hence, option (c).

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.

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:

1. ​​​​​​​N = 0 to 4 (no zeroes in N!) 
2. N = 5 to 9 (1 zero in N!) 
3. N = 10 to 14 (2 zeroes in N!) 
4. 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 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.

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.

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.