Question 1

Q1. An old man has Rs (1! + 2! + 3! + ...+ 50!), all of which he wants to divide equally (without fractions) among his n children. Then, n may be

(a)5

(b)7

(c)9

(d)11

Solution:

If the man is dividing the amount (1! + 2! + 3! + … + 50!) among n children, then this sum should be divisible by n.

Hence in this problem, we work with options.

Lets start with option (a). So, now, let us suppose that n = 5. Now we find the remainder when (1! + 2! + 3! + … + 50!) is divided by 5. If the remainder is 0, that means the amount (1! + 2! + 3! + … + 50!) is perfectly divisible by 5 and hence a possible value of n can be 5 and if we get any other non – zero remainder, then the amount is not divisible by 5 and hence n cannot be equal to 5 in that case.

Now, note that of the given terms, in the amount, all the terms from 5! onwards are divisible by 5 and hence we have to find the remainder when (1! + 2! + 3! + 4!) is divided by 5. Now we can check that 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 and this is not divisible by 5 and hence n cannot be equal to 5.

Next we move on to option (b). Let us suppose that n = 7. Note that of the given terms, in the amount, all the terms from 7! onwards are divisible by 7 and hence we have to find the remainder when (1! + 2! + 3! + 4! + 5! + 6!) is divided by 7. Now we can check that 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 and this is not divisible by 7 and hence n cannot be equal to 7.

Next we move on to option (c). Let us suppose that n = 9. Note that of the given terms, in the amount, all the terms from 6! onwards are divisible by 9 and hence we have to find the remainder when (1! + 2! + 3! + 4! + 5!) is divided by 9. Now we can check that 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 and this is divisible by 9 and hence n can be equal to 9.

Therefore the possible number of children can be 9.

Thank You.

Ravi Raja

Question 2

Q2.Find two whole numbers which, when multiplied together give an answer of 41.

Solution:

41, being a prime number, it can be expressed as a product of two numbers in exactly one way and that is 1 x 41. hence the two required whole numbers are 1 and 41.

Thank You.

Ravi Raja

Question 3

Q3. What is the last digit of the number you get by multiplying the first 2002 odd prime numbers together?

a. 1

b. 3

c. 5

d. 7

e. 9

Solution:

Since one of the prime numbers is 5 and the rest of them are all odd primes, hence the last digit of the product of these numbers has to be 5. Since any odd number when multiplied by 5, the last digit of the result will always be 5.

Thank You.

Ravi Raja

Question 4

Q4. Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product? (1) 1050 (2) 540 (3) 1440 (4) 1590 Solution: Let one of the numbers be x. He was supposed to find the product 35x but by mistake he calculated the product 53x and the new result was greater than what he was required to find by 540. That is, 53x – 35x = 540 or, 18x = 540 or, x = 540/18 = 30 Hence the final product was: 53x = (53)(30) = 1590. Thank You. Ravi Raja

Question 5

Q5.A tennis tournament is held in a school where every student plays 1 game against every other student. There are 210 boys vs. boys games held and 300 girls vs. girl games. Q19. How many boy vs. girl games were held? a. 40 b. 225 c. 525 d. 810 Solution: Let us suppose that the number of boys is x and the number of girls is y. Now, the first boy plays with the remaining (x – 1) boys, the second boy plays with the remaining (x – 2) boys, the third plays with (x – 3), … and so on. Hence the total number of boys vs. boys games will be: 1 + 2 + 3 + … + (x – 2) + (x – 1) = x(x – 1)/2 Similarly, the total number of girls vs. girl games will be y(y – 1)/2 Now, it is given that: x(x – 1)/2 = 210 or, x(x – 1) = 420 or, x^2 – x – 420 = 0 or, (x – 21)(x + 20) = 0 or, x = 21 (as x cannot be negative) and y(y – 1)/2 = 300 or, y(y – 1) = 600 or, y^2 – y – 600 = 0 or, (y – 25)(y + 24) = 0 or, y = 25 (as y cannot be negative) Now, if there are x boys and y girls, then the number of boys vs. girls games will be (x)(y) = (21)(25) = 525, which is the required answer. Thank You. Ravi Raja

Question 6

Q6. For how many values of k is 12^12 the LCM of the positive integers 6^6, 8^8 and k?

(a) 24

(b) 1

(c) 25

(d) 25 x 13

Solution:

6^6 = (2 x 3)^6 = (2^6) x (3^6)

8^8 = (2^3)^8 = 2^24

12^12 = {(2^2) x (3)}^12 = (2^24) x (3^12)

Now, we know that the L.C.M. contains the highest powers of the prime numbers involved in the prime factorization of the set of numbers. Hence, it is clear that the third number k, must contain 3^12 as one of its factors and it can contain 2 as its factor in (24 + 1) = 25 ways.

Hence there are 25 possible values of k and those possible values are: (3^12), (3^12) x (2^1), (3^12) x (2^1), (3^12) x (2^2), (3^12) x (2^3), ……, (3^12) x (2^22), (3^12) x (2^23), (3^12) x (2^24).

Thank You.

Ravi Raja

Question 7

Q7. A very fast growing sunflower grows to a height of 12 feet in 12 weeks by doubling its height every week. If you only want your sunflower to be 6 feet tall, after how many weeks should you stop it growing? Solution: Since the sunflower doubles its height every week and it was 12 feet in 12 weeks, then clearly in the previous week, its height was half of 12 feet. Hence if I want the sunflower to be 6 feet tall, then I should stop its growth at the end of 11 weeks. Thank You. Ravi Raja

Question 8

Q8. The sum of three consecutive numbers in a geometric sequence is 70. If the first is multiplied by 3, the second by 4 and the third by 4, the resulting numbers will be consecutive numbers in an arithmetic sequence. If each of the original numbers is a whole number, the first number is a. 30 b. 70 c. 15 d. 25 e. 40 Solution: Let the three numbers be a, ar and a(r^2). We know that: a + ar + a(r^2) = 70 or, a(1 + r + r^2) = 70 ------------- (1) We also know that the numbers: 3a, 4ar and 4a(r^2) are in A.P. That is, 2(4ar) = 3a + 4a(r^2) or, 8ar = 3a + 4a(r^2) or, a{3 – 8r + 4(r^2)} = 0 ------------- (2) So, wither a = 0 or 3 – 8r + 4(r^2) = 0 But a cannot be equal to 0. Therefore, 3 – 8r + 4(r^2) = 0 or, 4(r^2) – 6r – 2r + 3 = 0 or, (2r – 1)(2r – 3) = 0 or, r = 1/2 or 3/2 When r = 1/2, we get from equation (1), a = 40 When r = 3/2, we get from equation (1), a = 280/19 But it is given that the numbers are whole numbers. Hence the first number is 40. Thank You. Ravi Raja

Question 9

Q9. Let x, y and z be distinct integers x and y are odd and positive and z is even and positive. Which one of the following statements cannot be true?

(1) {(x – z)^2}(y) is even

(2) (x – z)(y^2) is odd

(3) (x – z)y is odd

(4) {(x – y)^2}(z) is even

Solution:

x and y being odd and z being even, we can check that (x – z) = odd – even = odd and hence (x – z)^2 will be odd and since (odd) x (odd) = odd, therefore, {9x – z)^2}(y) will always be odd and can never be even. So, option (a) is our required answer.

Thank You.

Ravi Raja

Question 10

Q10. Find the natural number below 1000, which has the maximum number of divisors. Find the sum of its digits?

a. 9

b. 18

c. 12

d. 15

Solution:

Check that 12 = 2 * 2 * 3 has 6 divisors: 1, 2, 3, 4, 6, and 12.

Restricting the primes to 2's and 3's, 2^5 * 3^3 = 864 and so we see that the number of divisors of 864 is 24.

Testing for numbers like n = 2^a * 3^b * 5^c we find that 2^4 * 3^2 * 5 = 720 and we see that the number of divisors of 720 = 30, also 2^6 * 3 * 5 = 960 and we see that the number of divisors of 960 is 28, but if we allow 7's then 2^3 * 3 * 5 * 7 = 840 and we see that the number of divisors of 840 is 32.

This is the answer: 840 has 32 divisors.

[Note: There is no definite method to find the greatest number of divisors for a number in a given range. All we have to do is use the trial method and restrict the number to less than 1000 (as given in the problem) and then calculate the number of divisors].

Thank You.

Ravi Raja

Question 11

Q11. (a + b + c + d + ...)^23 = a^23 + b^23 + c^23 + d^23 + ....+ M, then M is divisible by (a) 23 (b) 17 (c) 11 (d) Can’t be determined Solution: (a + b + c + d)^1 = a + b + c + d + [ 0 ] (Since there is no extra term, here, M = 0) (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + [2(ab + ac + ad + bc + bd + cd)] (a + b + c + d)^3 = a^3 + b^3 + c^3 + d^3 + [3ab(a + b) + 3cd(c + d) + 3(a + b)(c + d)] …………………… …………………… …………………… …………………… and so on. So, note that for any power n, the part of the expansion within the third bracket [ ] is denotes the M – part in the given problem and in each case, we note that M is divisible by n. Hence, in the given problem, n = 23 and so we can conclude that M is divisible by 23. [Note: Here I have solved the problem by seeing the pattern that is followed for any power ‘n’. We can also use the result of Binomial Expansion and check that M will be divisible by 23]. Thank You. Ravi Raja

Question 12

Q12. Place an ordinary mathematical symbol between 2 and 3 so that the result is a number which greater than 2 and less than 3. Solution: Insert a decimal sign between 2 and 3 to get the result 2.3 which is greater than 2 and less than 3. Thank You. Ravi Raja

Q 10 can we not solve like

can we not solve like this.............

here for max. no. of divisors of a number ,we have to find the max no. of prime no. which can be inculcated in the given range.

here 2,3,5,7 can be inculcated as there product is 210 .we can not involve 11 as it will cause the product to be greater than 1000.

now,210x1=210

210x2=420

210x3=620

210x4=840

210x5=1050 which is greater than 1000.

so answer comes out to be 840. with in seconds.

ashish kumar

  Q11 here if we take this

Q11

here if we take this general Q. for only two variables a and b then it can be solved very quickly.

(a+b)^23 can be elaborated by binomial terms .then we will find that except a^23 and b^23 all terms are divisible by 23 . hence 23 is the answer.

ashish kumar

hello ravi

hello ravi raja,

       My name is Shahid. I'm a gazetted officer of andhra pradesh government. I want to do MBA and that too from

a reputed institution like IIM-A. For that i'm preparing for CAT. Since i don't have time to attend any coaching

classe i'm preparing at home. I have collected some good books and to procure any additional material and to be

in touch with other MBA aspirants i got a broadband connection. To tell the truth i'm not well verse with intenet

and it took me nearly 3 months to findout this website. I have seen solutions given by you to some of tricky

questions. They are very methodical and quite easily understandable. Your solutions helped me a lot and i

sincerely thank you for that.

     Keep the good work and best of luck,

                                                                                                                         yours thankfully,

                                                                                                                              shahid.

A serious doubt...

I do agree with ur solution to this problem. But when i solved this taking the first three numbers as a(r^-1),a and a(r^1) then a(r^-1)+a+ar=70 and 8a=12(a^2) which gives a =2/3. Can u justify, what was the mistake. I went that way in choosing the numbers because in AP or GP its advised to take 3 numbers as a-d, a and a+d 4 nubers as a-3d, a-d, a+d and a+3d... similar way in gp..

and 8a=12(a^2) which gives a

and 8a=12(a^2)
which gives a =2/3.

 3a(r^-1),4a and 4a(r¹)

4ar - 4a = 4a - 3a(r^-1)

=>3(r^-1) = 8 - 4r

=> 4r + 3(r^-1) = 8

=>4r² -8r + 3 = 0

=>4r² - 6r - 2r + 3 = 0

=>2r (2r - 3) -1 (2r - 3) =0

=>r = 1/2 or r = 3/2

Now a(r^-1)+a+ar=70

2a + a + 1/2 a = 70

=>7a/2 = 70

=>a = 20

First number = 40

and the sequqnece is 40, 20, 10

ans 4

53x-35x=540
therfore x=30
hence the new product is 53*30=1590

we can also put the

we can also put the mathematical symbol between 2 and 3 as squareroot , i.e 2sqrt(3)

the solution for this

the solution for this problem may be calculated from the answers. take option 1)1050, add 540 to it you will get 1590 which is divisible by 53.