Tag: some special sequences

Questions Related to some special sequences

If the square of a number ends with $10$ zeroes, how many zeroes will the number have at the end?

  1. $5$

  2. $15$

  3. $30$

  4. $40$


Correct Option: A
Explanation:

If the square of a number has $ 10 $ zeroes, the number will have $ 10 \div 2 = 5 $ zeroes.

If a number ends with 3 zeroes, how many zeroes will its square have at the end ?

  1. 3

  2. 4

  3. 6

  4. 1


Correct Option: C
Explanation:

If a number ends with $ 3 $ zeroes, then its square will have $ 3 \times 2 = 6 $ zeroes.

Express $49$ as the sum of $7$ odd numbers.
Express $121$ as the sum of $11$ odd numbers.

  1. $1+3+5+7+9+11$
    $1+3+5+7+9+11+13+15+19$

  2. $1+3+5+7+9+11+13$
    $1+3+5+7+9+11+13+15+19+21$

  3. $1+3+5+7+9+11$
    $1+3+5+7+9+11+13+15+19+21$

  4. $1+3+5+7+9+11+13$
    $1+3+5+7+9+11+13+15+19$


Correct Option: B
Explanation:

$49=7^2=$ sum of first $7$ odd numbers.
So, $49=1+3+5+7+9+11+13$.
Similarly, $121=11^2=$ sum of first $11$ odd numbers. 
So, $121=1+3+5+7+9+11+13+15+19+21$

Observe the following pattern and fill in the missing number. 
$ \displaystyle 11^{2} =121$
$ \displaystyle 101^{2} =10201$
$ \displaystyle 10101^{2} =102030201$
$ \displaystyle 1010101^{2} =......................$

  1. $ \displaystyle 1010101^{2} $=10203030201

  2. $ \displaystyle 1010101^{2} $=10204040201

  3. $ \displaystyle 1010101^{2} $=1020304030201

  4. $ \displaystyle 1010101^{2} $=10204030201


Correct Option: C
Explanation:

$11^{2}$$=$$121$

$101^{2}$$=$$10201$
$10101^{2}$$=$$10203020101$
$1010101^{2}$$=$$1020304030201$
$101010101^{2}$$=$$10203040504030201$
We will go up to number of  ones in the number numerically.
Hence, Option C is correct.

State whether true or false:

121 can be expressed as the sum of 11 odd numbers.

  1. True

  2. False


Correct Option: A
Explanation:

$\sum _1^n(2n+1)=n^2$

Sum of  n odd consecutive number$=n^2$
$1+3+5+7+9+11+13+15+17+19+21=(11)^2=121$  (as there are 11  odd numbers n=11)

Find the sum of the following odd numbers given .

$1+3+5+7+9+11+13$

  1. $25$

  2. $36$

  3. $49$

  4. $100$


Correct Option: C
Explanation:

Sum of odd numbers is $n^2$ where n is no. of odd nos.
n= 7 odd nos. 
Therefore sum is $7^2  = 49$

Find the value of the following without actually multiplying:

 $13 \times 15$

  1. 165

  2. 145

  3. 156

  4. 195


Correct Option: D
Explanation:

$13\times15$ $=$ $(10+3)$ $(10+5)$

$100+50+30+15$ $=$ $195$
Hence, Option D is correct.

Find the sum of the following series without actually adding it.

$1+3+5+7+9+11+13+15+17+19+21$

  1. $101$

  2. $161$

  3. $121$

  4. $141$


Correct Option: C
Explanation:

Given series is $1+3+5+7+9+11+13+15+17+21$

Sum of odd natural numbers is $=n^2$

where $n= $ number of odd numbers are $11$

So, sum of the series given is $=11^2= 121 $.

State whether true or false:

When an even number is given, square of this number will be even.

  1. True

  2. False


Correct Option: A
Explanation:

$28 \times  28 = 78\underline{4} $
$162 \times  162 = 262\underline{44}$
(last digit is 4 even , so whole number is even.)

Therefore, A is the correct answer.

Find the value of $7 \times 9$.

  1. 64

  2. 63

  3. 53

  4. None of these


Correct Option: B
Explanation:

$7\times9$ $=$ $(10-3)$ $(10-1)$

$100-30-10+3$ $=$ $63$
Hence, Option B is correct.