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

12. In a certain examination paper there are n questions. For j=1,2,3,.....n, there are 2^(n-1) students who answered 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

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

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;

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.

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