eLitmus Number System Questions Quiz 1(A)

Solution:

Since x should be greater than 121

Let us suppose that 'y' is no. which is greater than 0,

For this we can write, x^2+1 as (121+y)^2 +1.

(y^2 + 121^2 + 2*121*y + 1)/11

y^2/11 + 121^2/11 + 2*121*y/11 + 1/11

y^2/11 + 0 + 0 + 1 ( remainder)

y^2/11 + 1

y^2 can be 1,4,9,16,25..

Now we will divide this by 11 and the remainder we get it ,4,6 but not 8.

So the correct answer to this question is option D: 8

 

First we fix 1 at unit place of a six digit number.

Since any number can be divided by 1 .

So it does not matter what comes at other five places .
_ _ _ _ _ 1
So, first position can be filled in 5 ways as any number
from 2 to 6 can be placed in that position .
Similarly 2nd position can be filled with 4 ways .

So , by fundamental principle of counting ,
We get,

5x4x3x2x1x1=120 or 5! x 1=120

Now for 2 at units place ,

Any even number is divided by 2 ,
Hence it is again similar to first case,
5! × 1 =120

Now for 3 at units place ,

For any number to be divisible by 3 , we need to make
sure that sum of the digits is divisible by 3.

Now , 1+ 2 + 3 +4 + 5 +6 = 21
Since 21 is divisible by 3 , so any number formed by number
1 to 6 is divisible by 3 .
Hence , 5! × 1 =120.

Similarly for 5 , 5! ×1 = 120

Now for 6 at unit place ,
We know that any number which is divisible by 2 and
3 is also divisible by 6 .

Hence , 5! ×1 =120

Now for 4 at unit place.
If there is even number at tens place with 4 at units place ,
then the number is divided by 4 else no.

So we have only two even number 2 and 6 which we can
put at tens place .
So now first fixing 2 at tens place and 4 at units place .
_ _ _ _ 2 4

We get by fundamental principle of counting ,
4 × 3 × 2 × 1 × 1 ×1
Or
4! × 1 ×1 = 24

Similarly for 6 at tens place and 4 at units place .
We get 4! × 1 × 1 = 24

Now let us add all the values we got ,

120 +120 +120 + 120 +120 +24 +24 = 648

Hence answer is (c)

N= 1!+2!+3!.....+10!
Summing up the factorials gives,
N=1+2+6+24+120+720+5040+40320+362880+3628800=4037913
Now we have to find the last digit of N^N
N^N=(4037913)^4037913
for finding out the last digit, we can write only the last digit of
4037913 which is 3, hence
(3)^4037913
[(3)^(4*1009478)]*3 ............using cyclicity of 3
1*3=3 ans

because the result trailing 0 occurs when one 2 and one 5 multiplied.

here we see the set M:

( 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 35, 45, 55, 65, 75, 85, 95 )

after 22 I have taken only the term which contains the factor of 5

now we find the prime factor

2 = 2
4 = 2 * 2
6 = 2 * 3
8 = 2 * 2 * 2
10 = 2 * 5
12 = 2 * 2 * 3
14 = 2 * 7
16 = 2 * 2 * 2 * 2
18 = 2 * 3 * 3
20 = 2 * 2 * 5
22 = 2 * 11
25 = 5 * 5
35 = 5 * 7
45 = 5 * 9
55 = 5 * 11
65 = 5 * 13
75 = 5 * 5 * 3
85 = 5 * 17
95 = 5 * 19

here we find
total number of 5 is 12
and
total number of 2 is 19

and
we get the total number of 5 * 2 pair is 12
so the number of trailing zero will be 12

ans is
d)
because e and 0 can\\\\\\\\\\\\\\\'t be more than w, so in this case only one option left. by the way option a and b are same.

9 div 6=3 rem

3*11=33
33/6 =3 rem

3*22=66
66/6= 0rem

99/6= 3 rem

so 3 and 0 ill be rem( when even no. comes rem will be 0 else 3)

Odd*odd=odd
Odd+odd=even
Even*even=even
Each and every even no. Is divisible by 2 it is not necessary
that every even no. Is divisible by 3,5 or 10 but every
even no get divide by 2
So ans is 2...

s(n) = sum of n natural numbers = n(n+1)/2 because
n numbers of natural numbers are 1+2+3+4+5+...+n which are in a.p. with d=1 a= 1 and LAst Term(L) = n
so by formula s(n ) = n/2*(a+L)
s(n) = n(n+1)/2

putting this in equation
not going to show calculation
z(n) = n(n+1) +41

some times after some steps u need to hit and try in elitmus cause that is the faster way
so for n= 41 we can use reverse substitution and by that i mean instead of 41 put n
z(n)= n(n+1) + n
z(n) = n(n+2)
by this we can see z(n) is not prime as it is a multiplication of two numbers a prime number can only have factors 1

for 41 z(n) = 1681
which has factors 1,41 and 1681 so
answer is (d)