# Quant Questions With Solutions- 2

11. If v,w,x,y,z are non negative integers each less than 11, then how many distinct combinations are possible of(v,w,x,y,z) which satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z = 151001 ?

Changing 151001 to base 11 number we get,

it will be

A34A4 i.e.

10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4

it will be

A34A4 i.e.

10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4

12. In a certain examination paper there are n questions. For j=1,2,3,.....n, there are 2^(n-1) students whoanswered j or more question wrongly. If the total number of wrong answers is 4096 then the value of n is

a)13

b)11

c)10

d)9

Given that,

2^(n-1) = 4096 = 2^12

i.e)n-1=12

==> n=13

2^(n-1) = 4096 = 2^12

i.e)n-1=12

==> n=13

13. How many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at it's unit's place?

As each number will contain all the six digits and the sum of digits is = 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3.

The numbers ending with digit 1 will be divisible 1.

The numbers ending with digit 2 will be divisible 2.

The numbers ending with digit 3 are divisible 3.

The numbers ending with digit 5 will be divisible 5.

The numbers ending with digit 6 will be divisible 6.

Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.

The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)

so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648

The numbers ending with digit 1 will be divisible 1.

The numbers ending with digit 2 will be divisible 2.

The numbers ending with digit 3 are divisible 3.

The numbers ending with digit 5 will be divisible 5.

The numbers ending with digit 6 will be divisible 6.

Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.

The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)

so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648

14. A natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have?

Consider the no is 512, the divisors of this no. are

1,2,4,8,16,32,64,128,256,512;

so only one prime factors are there i.e. 2;

if the no is 48 the divisors are

1,2,3,4,6,8,12,16,24,48;

so the prime factors are 2,3;

1,2,4,8,16,32,64,128,256,512;

so only one prime factors are there i.e. 2;

if the no is 48 the divisors are

1,2,3,4,6,8,12,16,24,48;

so the prime factors are 2,3;

15. If m and n are two positive integers, then what is the value of mn? Given:

(1)7m + 5n= 29

(2) m + n= 5

7m + 5n = 29

m + n = 5(multiply both sides by 7)

subtract equation second from first,we get

7m + 5n =29

7m + 7n = 35

n = 3

substitute value of n in any equation.

m = 2

mn = 6.

m + n = 5(multiply both sides by 7)

subtract equation second from first,we get

7m + 5n =29

7m + 7n = 35

n = 3

substitute value of n in any equation.

m = 2

mn = 6.

16. A natural number has exactly 10 divisors including 1 and itself. How many distinct prime factors can this natural number have?

A. Either 1 or 2

B. Either 1 or 3

C. Either 2 or 3

D. Either 1,2 or 3

**Solution:**

**Ans) Either 1 or 2**

Check on 2

^{9}i.e 512 , 3^{9}, 5^{9 }which have only 1 prime factor and 80 , 48 which are having 2
prime factors and total of 10 divisors.

Let us also consider the case of 3 prime factors. Let x, y ,z be the three prime factors of a

number. Therefore 1 , x ,y ,z ,xy ,yx, zx ,xyz must be the factors of that nos . We have

minimum 8 such factors with xyz as the nos or the factor of the nos.

When xyz is the nos then we will have exactly 8 divisors but if the nos is greater thn xyz that

is a multiple of xyz , either the nos is multiplied by any of these prime factors x , y , z then

we will get at least 12 divisors. So we don’t get 3 prime factors with 10 divisors.

17. What is the remainder when 128^1000 is divided by 153?

128^1000 = (153-25)^1000 = (25^1000)mod153 = (625^500)mod153 = [(4*153+13)^500]mod153 = (13^500)mod153 = (169^250)mod153 = [(153+16)^250]mod153 = (16^250)mod153 = (256^125)mod153 = [(153+103)^125]mod153 = (103^125)mod153 = [(2*3*17+1)^125]mod153

At this point,observe that 153=17*(3^2);

Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.

Actually,the above line can be written since only except the last two

terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,

[153*68] as one of its factors.

Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;

[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);

which means [125*2*3*17 + 1] = 83*153 + 52;

which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ;

So, remainder is 52.

At this point,observe that 153=17*(3^2);

Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.

Actually,the above line can be written since only except the last two

terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,

[153*68] as one of its factors.

Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;

[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);

which means [125*2*3*17 + 1] = 83*153 + 52;

which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ;

So, remainder is 52.

18. Given a,b,c are in GP and a < b < c. How many different different values of a, b, c satisfy (log(a) + log(b) + log(c) ) = 6?

log(abc)=6

abc=10^6 because of b^2=ac=100,

b^3=10^6,

b=10^2=100

and the combinations are

(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).

abc=10^6 because of b^2=ac=100,

b^3=10^6,

b=10^2=100

and the combinations are

(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).

19. What will be the remainder when expression 2^2+22^2+222^2+2222^2+....+22222...48times^2 is divided by 9?

First, let us consider a general case :

(222222222222.....{2 is repeated n times})^2

=[2(111111111111........{1 is repeated n times})]^2

=4(111111111111........{1 is repeated n times})^2

=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]

Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);

Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].

Now, looking at the [(eqn1)] ,

we can find that here k =(n-1), (n-2), (n-3),....,1.

Now,for 10^(n-1) = 9*(a1) + 1;

10^(n-2) = 9*(a2) + 1;

10^(n-3) = 9*(a3) + 1;

10^(n-4) = 9*(a4) + 1;

.

.

.

.

10^1 =9 +1;

1 = 1

So,now in [(eqn1)] , we can write

(222222222222.....{2 is repeated n times})^2

=4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2

=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.

=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]

Now this a general expression for (222222222222.....{2 is repeated n times})^2.

For the given problem,we can find that n=1,2,3,4,5,6,....48.

In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have

to consider the last term,i.e, 4*(n^2);

Summing up [4*(n^2)] for n=1,2,...48. we get

4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}

=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]

=4*8*49*97

=4(9- 1)(9*5 +4)(9*11 -2)

=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important

=9*B + 32

=9*B + 9*3 + 5;

So, clearly the remainder will be 5.

(222222222222.....{2 is repeated n times})^2

=[2(111111111111........{1 is repeated n times})]^2

=4(111111111111........{1 is repeated n times})^2

=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]

Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);

Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].

Now, looking at the [(eqn1)] ,

we can find that here k =(n-1), (n-2), (n-3),....,1.

Now,for 10^(n-1) = 9*(a1) + 1;

10^(n-2) = 9*(a2) + 1;

10^(n-3) = 9*(a3) + 1;

10^(n-4) = 9*(a4) + 1;

.

.

.

.

10^1 =9 +1;

1 = 1

So,now in [(eqn1)] , we can write

(222222222222.....{2 is repeated n times})^2

=4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2

=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.

=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]

Now this a general expression for (222222222222.....{2 is repeated n times})^2.

For the given problem,we can find that n=1,2,3,4,5,6,....48.

In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have

to consider the last term,i.e, 4*(n^2);

Summing up [4*(n^2)] for n=1,2,...48. we get

4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}

=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]

=4*8*49*97

=4(9- 1)(9*5 +4)(9*11 -2)

=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important

=9*B + 32

=9*B + 9*3 + 5;

So, clearly the remainder will be 5.

20. Given that a number Q < 200, calculate sum of all Q such that when Q divided by 5 or 7 givesremainder 2?

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in question 13

ReplyDeleteThe no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)

how this step is caluclated ....and how did we get 72 there ?

can anyone explain how did we get solution for question 11 ? please

ReplyDelete3c1 * 4!

ReplyDelete