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)

According to the question 2^(n-1) students got 1 or more questions wrong.

2^(n-2) students got 2 or more questions wrong.

2^(n-3) students got 3 or more questions wrong.. and so on...

So we can tell that 2^(n-1) – 2^(n-2) students got exactly one question wrong.

2^(n-2) – 2^(n-3) students got exactly two question wrong.

2^(n-3) – 2^(n-4) students got exactly three question wrong and so on..

Now to compute the total number of wrong answers equation would be:

1 * [2^(n-1) – 2^(n-2)] + 2 * [2^(n-2) – 2^(n-3)] + 3 * [2^(n-3) – 2^(n-4)] +…… + (n-1)[2 - n^0] + n

= 2^(n-1) + 2^(n-2) + 2^(n-3) + 2^(n-4) + ........ + 2 - n + 1 + n

Now clearly 2^(n-1) + 2^(n-2) + 2^(n-3) + 2^(n-4) + ........ + 2 + 1 is a GP where a = 1, r = 2

So, 1(2^n – 1) / (2 – 1)
= 2^n - 1, which is the total number of wrong answers.

Acc. to question, Total No. of wrong answered question is 4095.
Therefore,
2^n - 1 = 4095
=>n = 12. Ans

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.

9^1+(9^2+9^3+9^4+9^5+9^6+9^7+9^8+9^9). Notice that in the brackets we have the sum of 8 odd multiples of 3, hence the sum in the brackets will be even multiple of 3 (the sum of 8 odd numbers is even). So, the sum in the brackets is multiple of 6 (remainder is zero). So we are just left with the first term 9, which yields remainder of 3 upon division by 6.

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) = 1763
which has factors 1,41 and 1763 so
answer is (d)

Let the number  be 476ab0

 

476ab0 is divisible by 3

=> 4 + 7 + 6 + a + b + 0 is divisible by 3

=> 17 + a + b is divisible by 3 ------------------------(i)

 

476ab0 is divisible by 11

[(4 + 6 + b) -(7 + a + 0)] is 0 or divisible by 11

=> [3 + (b - a)] is 0 or divisible by 11  --------------(ii)

 

Substitute the values of a and b with the values given in the choices and select the values which satisfies both Equation 1 and Equation 2.

 

if a=6 and b=2,

17 + a + b = 17 + 6 + 2 = 25 which is not divisible by 3 --- Does not meet equation(i).Hence this is not the answer

 

if a=8 and b=2,

17 + a + b = 17 + 8 + 2 = 27 which is divisible by 3 --- Meet equation(i)

[3 + (b - a)] = [3 + (2 - 8)] = -3 which is neither 0 nor divisible by 11---Does not meet equation(ii).Hence this is not the answer

 

if a=6 and b=5,

17 + a + b = 17 + 6 + 5 = 28 which is not divisible by 3 --- Does not meet equation (i) .Hence this is not the answer

 

 

if a=8 and b=5,

17 + a + b = 17 + 8 + 5 = 30 which is divisible by 3 --- Meet equation 1

[3 + (b - a)] = [3 + (5 - 8)] = 0 ---Meet equation 2

Since these values satisfies both equation 1 and equation 2, this is the answer