This is the post in the CAT 2016 Sprint Preparation Series – Numbers. We have posted 20 questions from previous year CAT papers, forums, mock tests, and other entrances that are on par with the level of difficulty you can expect in CAT 2016. We will be posting the solutions and traps/things by the end of the day to look at while solving similar questions so that you are avoid making silly mistakes during the test.

You can go through the entire series by clicking on this link: CAT 2016 Sprint Preparation Series by Learningroots.

Quant | Set 2 | Numbers

CAT 2016 Sprint Preparation Series

 

1. The sum of thirty-two consecutive natural numbers is a perfect square. What is the least possible sum of the smallest and the largest of the thirty-two numbers?

2. In a three-digit number, the unit digit is twice the tens digit and the tens digit is twice the hundreds digit. The same number is written as 1XY and 1YX in base 8 and base 9 respectively. Find the sum of X and Y in the decimal system.

3. What are the last two digits of 5626^1347?

4. How many numbers are there between 0 and 500 which when divided by 3, 4, 6 and 8 leave remainders 1, 2, 4 and 6 respectively?

a. 10

b. 20

c. 21

d. None of these

5. The H.C.F. of a, b and c is 8. If a – b = b – c = 8 and the L.C.M. of a, b and c is a four-digit number, then what is the maximum possible value of c?

a. 80

b. 88

c. 96

d. 100

6. How many divisors of 25200 can be expressed in the form 4k + 3, where k is a whole number?

7. The HCF of three natural numbers x, y and z is 13. If the sum of x, y and z is 117, then how many ordered triplets (x, y, z) exist?

8. The digits of a 3-digit number in Base 4 get reversed when it is converted into Base 3. How many such numbers exist?

9. How many 4-digit multiples of 3 can be formed using the digits 2 and 3 only?

10. A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1’s, then six consecutive 2’s, and likewise six 3’s, six 4’s and six 7’s in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.

11. If ‘c’ is a positive integer, how many three-digit numbers ‘abc’ are there such that both ‘abc’ and ‘cba’ are divisible by 4?

12. For how many of the first 300 even natural numbers is the total number of factors even?

a. 88

b. 276

c. 288

d. 90

13. What is the sum of all the five-digit numbers that can be formed by arranging the digits 2, 3, 4, 5 and 6, using each digit once and only once?

14. On January 1, 2004 two new societies s1 and s2 are formed, each n numbers. On the first day of each subsequent month, s1 adds b members while s2 multiples its current numbers by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r?

15. Suppose n is an integer such that the sum of digits of n is 2, and 10^10 < n < 10^11. The number of different values of n is ?

16. The remainder, when 98^3 + 99^3 + 101^3 + 102^3 is divided by 400, is

17. The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

a. 100 < A < 299

b. 106 < A < 305

c. 112 < A < 311

d. 118< A < 317

18. For a positive integer n, let pn denote the product of the digits of n and sn denote the sum of the digits of n. The number of integers between 10 and 1000 for which pn + sn = n is

19. Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is

20. Consider four-digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

 

Solutions:

 

1. Whenever in doubt, start by taking a variable. Let the first integer in the series be x. So, the rest of the series will be: x, x + 1, x + 2 … x + 31

Sum of the series will be:

32x + 31*16

= 32x + 496

= 16 (2x + 31)

Now, this has to be a perfect square and so, if one part of it is a perfect square, the other part should also be a perfect square. So, 2x + 31 is a perfect square. As x is greater than 1 and 2x + 31 is an odd number, we have to find the next odd number which is also a perfect square i.e. 49. So, the first number will be x = 9 and the last number will be x + 31 = 40. So, the sum will be 49.

2. Again, it might look confusing at the onset as people are generally not comfortable dealing with multiple bases. So, what we can do is to convert the bases to decimal and then see how it goes.

1XY in base 8 corresponds to 64 + 8X + Y

1YX in base 9 corresponds to 81 + 9Y + X

As both these numbers are the same in the decimal notation, we get

64 + 8X + Y = 81 + 9Y + X

7X – 8Y = 17

The smallest value of X that satisfies this equation is 7 and so, the next such value will be 15 which is outside the purview of the question. So, the number in base 8 will be 174 and so, X + Y will be 11 in the decimal system.

PS: The first part of the question is practically useless.

3. It would be an easy question for someone who is fluent with the concept. There are a lot of ways of doing this. Probably the most non-invasive one is:

26^1 = 26

26^2 = 76

26^3 = 76

26^4 = 76 and so on. So, every power of 76 ends in 76. So that is our answer.

How many of you actually tried out Euler’s and the last 2 digits concept?

4. Another typical multiple divisor/dividend question.

Using the concept of negative remainders, we get that:

3a – 2 = 4b – 2 = 6c – 2 = 8d – 2 = N

So,

N + 2 = 3a = 4b = 6c = 8d

So, the smallest such number will be the LCM of 3, 4, 6 and 8 i.e. 24.

So, the number will be in the form of 24k – 2

The smallest value of k which is possible is 1 which gives us 22. To find the largest such value, we have to solve the smallest solution to this inequality

24k – 2 > 500

24k > 502

k > 20.xx

So, the smallest value of k for which the number goes beyond 500 is 21. So, for our solution, the largest value of k that satisfies will be 20 which will give us 478 as the largest value.

Total of 20 numbers.

5. From the given data, we get that a = b + 8 and b = c + 8. So, these are three numbers separated by 8 each such that a > b > c.

Let c be equal to 8x, then b will be equal to 8x + 8 and a will be equal to 8x + 16.

LCM of the three numbers will be 8x(x + 1)(x + 2) if x is odd and will be 16(x/2)(x + 1)(x + 2)/2 or 4x(x + 1)(x + 2). It is obvious that we will the maximum value of c in the second case (because, if the coefficient is lesser, the value of x will be more) and so, let’s see what we get:

4x(x + 1)(x + 2) < 10000

x(x + 1)(x + 2) < 2500

x = 12 and so, maximum value of c will be 8x = 96.

6. 25200 can be written as 2^4 * 3^2 * 5^2 * 7^1

If the number is in the form of 4k + 3, it means that it should be odd in nature and should leave a remainder of 3 or -1 when divided by 4.

So, all the powers of 2 are out. If we look at the remaining part and divide it by 4, we can see that it is in the form of:

(-1)^2 * 1^2 * (-1)^1

Now, we want the resultant to be -1 and so, we will see the number of ways in which this can be done. The power of 1 doesn’t matter and so, it can be either 0, 1 or 2.

Now, the power of the first -1 can be either 0 or 2 if the power of the second -1 is 1 and the power of the first -1 could be 1 if the power of the second -1 is 0. So, we have a total of 3 cases. Considering the power of 1 as well, we get 3*3=9 cases in total.

7. Let the 3 numbers be 13a, 13b and 13c where a, b and c are coprime. So,

13a + 13b + 13c = 117

a + b + c = 9

As they are all natural numbers, we want the number of positive integral solutions:

a + b + c = 6

Total solutions are 8c2 = 28. Out of these, the only case that will be eliminated will be the one where all 3 are equal (and so, their HCF would no longer be 13). So, 27 solutions in total.

8. Let the 3 digit number be abc in base 4. So, it will be cba in base 3. Converting both these numbers to decimal and equating them, we get

16a + 4b + c = 9c + 3b + a

15a + b = 8c

Now, a, b and c have to be less than 3. The only case when this is possible is when a = b = 1 and c = 2.

9. For a number to be a multiple of 3, the sum of digits has to be a multiple of 3. The total number of ways in which we can form 4 digit numbers using 2 and 3 is 2^4 = 16. Least sum that is possible is 8 and the highest sum is 12. So, we consider the cases wherein the sum of the digits will be 9 or 12.

Sum of digits is 9 means that the digits will be 2, 2, 2 and 3. So, there are 4!/3! = 4 such numbers

Sum of digits is 12 means that the digits will be 3, 3, 3 and 3. So, there is just 1 such number.

Total of 5 numbers.

10. The initial number will be 111111222222…444444777777. We can see that this number is divisible by 11. The new number will be 111111222222…444444abc777 where abc is the three digit number that is inserted. Now, abc also has to be a multiple of 11 and so, we need to find all 3 digit numbers that are multiples of 11. So, all multiples from 110 to 990 i.e. 81 cases.

11. Instead of going for a formula based approach or dealing with a lot of variables simple observation would help. The three digit numbers that are multiples of 4 are 104, 108, 112, 116, 120, 124 …

So, if b is odd, a and c can be either 2 or 6. Total of 5*2*2 = 20 ways.

If b is even, a and c will have to be either 4 or 8. Total of 5*2*2 = 20 ways.

40 numbers in total.

12. The total number of factors is odd when the number is a perfect square (as it will have a mirror set of factors above and below the square root which means we will be left with one number at the end)

So, we have to calculate the number of even numbers from 2 till 600 that are perfect squares. Simple observation tells us that we have to count the number of even numbers from 2 to 24 which is 12 and so, the remaining 288 will have an even number of factors.

13. The direct rule is to calculate 11111*20*24 = 5333280.

14. n + 6b = n*r^6

As b = 10.5n, we get

r^6 = 2^6 and so, r = 2.

This one looked complicated but is extremely straightforward if you stick to it.

15. Basically we have to find the number of 11-digit integers whose sum of digits is 2. So, either there will be two 1s or one 2 in the number.

Two 1s: The first digit is 1 so the next 1 can be put in 10 different places.

One 2: The first digit is 2 so there will be just the one case.

Total of 11 cases.

16. This is a nice question to test your basics. Almost all the excited candidates will fall for the (a^3 + b^3 + c^3 + d^3) divided by (a + b + c + d) trap. Sample this: Let a, b, c and d be 1, 2, 4 and 5. The remainder is not 0 in this case, right? This is because there is an additional condition that a, b, c and d need to be in an arithmetic progression for the remainder to be 0.

98^3 + 99^3 + 101^3 + 102^3

(98^3 + 102^3) + (99^3 + 101^3)

200 * even number + 200 * odd number

200 * odd number

If this is divided by 400, we can cancel out 200 from the numerator and the denominator.

So, we are left with an odd number divided by 2 which will give the remainder as 1. Multiplying it back by the canceled factor 200, we get the final remainder as 200.

17. Let the original number be abc.

100a + 10b + c is the original number

100c + 10b + a is the reversed number

If we take the difference,

99c – 99a will be the resultant.

But, it is given that this is divisible by 7.

99(c – a) = 7k

So, c – a has to be a multiple of 7. So, it would be either 0 or 7. If c – a = 0, the two numbers will be equal which is not possible. So, c – a = 7.

The original number could be either 1b8 or 2b9. So, all numbers from 108 to 299 would be included.

Option b.

18. We have to find all two digit and three digit numbers that satisfy the criterion.

Let the two digit number be 10x + y

So,

xy + x + y = 10x + y

xy = 9x or y = 9 and x is not equal to 0.

So, all two digit numbers ending in a 9 will satisfy the criterion. Total of 9 cases (19, 29, … 99)

Let the three digit number be 100x + 10y + z

So,

xyz + x + y + z = 100x + 10y + z

xyz = 99x + 9y

yz = 99 +9(y/x)… dividing throughout by x which is not equal to 0.

The maximum possible value of yz will be 81 and so, the above equation will never be true.

So, total of 9 cases.

19. The good part is, the answer can be easily obtained. The bad part is, there is probably no better way to do it than by trial and error.

Two 70*30 tiles on one end of the 130 cm side and two 70*30 tiles on the other end of the 130 cm side would cover 120 cm horizontally. Also, there will be enough space to place another couple of tiles perpendicular to the original set of tiles. So, 6 in total. If you didn’t get this, shoot me a comment and I will put up the figure.

20. The classic CAT question that had a fact based answer for those who were aware. 7744 is the only number that satisfies the given condition. What if you didn’t know this?

You have to recollect that all squares will follow a mirrored cyclicity with regard to the last two digits around multiples of 50. It essentially means that 18^2 i.e. (0 + 18)^2 will have the same last two digits as 32^2 i.e. (50 – 18)^2 which will have the same last two digits as 68^2 i.e. (50 + 18)^2 and so on. If we look at the first 25 squares, we get that there are just two cases wherein the last two digits repeat themselves viz. 00 and 44. Now, we do not have a number of the form aa00 that is a perfect square and so, we go for the next case of aa44. Applying the abovementioned concept, we get that we have to figure out 12^2, 38^2, 62^2 and 88^2 for our solution. It will be obvious that only 88^2 = 7744 satisfies this.