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Common Information

$P =1234567891011.............9989991000$

Q.

Common Information Question: 1/2

Find the remainder when first 1000 digits of the number $P$ are divided by 16?

 A.

7

 B.

9

 C.

11

 D.

13

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Solution:
Option(D) is correct

$P$ is obtained by writing the first 1000 natural numbers from left to right.

We need last four digits of the number formed by the first thousand digits to check the divisibility of 16.

The first 99 natural numbers wll give us $9+(90×2) = 189$ digits.

We need another $1000-189 = 811$ digits from the remaining numbers numbers.
⇒ required additional natural numbers $=\dfrac{811}{3} = 270$ (approx)
i.e. $99+270 +\text{ one digit}$
i.e. first $\textbf{ 369}$ natural numbers and one additional digit will form a number containing the first $1000$ digits of the number $P$
i.e. end of the required number will look like
.........................$368\textbf{3693}$

⇒ remainder will be same as the remainder of $\dfrac{3693}{16} = \textbf{13}$


(7) Comment(s)


Shreyash Okhade
 ()

please tell me how to approach and solve these kind of question



PRATYUSH ANAND
 ()

sorry there is mistake in

iii)If you analyse carefully in 100101102....999

from 100 to 199 , 1 in hundred place is used 10 times it should be 100 times



PRATYUSH ANAND
 ()

iv) Sum of all digit in 1000 is 1

Now adding (i),(ii),(iii)&(iv),result to get sum of digits of 123456789..9979989991000 is=45+855+12600+1

=12501

And at the last if you divide (12501/16) you will get 13 as remainder

Hence answer is 13.



PRATYUSH ANAND
 ()

iii)If you analyse carefully in 100101102....999

from 100 to 199 , 1 in hundred place is used 10 times and 10 to 99 in tense place is appeared(100101102.....199)along with 1 (in hundred place)

and this is same for

200201202.....299

300301302.....399
.....
.....
.....
900901902.....999

Thus sum of

100101102....199=(1x100 +(sum of digits 0102030405...99(i.e same as above we did in (i for 123..9)&(ii for 10111213..99)and got 900 as result)))

=(100 + 900)

200201202....299=(2x100 +(sum of digits 0102030405...99(i.e same as above we did in (i for 123..9)&(ii for 10111213..99)and got 900 as result)))

=(200 + 900)

300301302....399=(3x100 +(sum of digits 0102030405...99(i.e same as above we did in (i for 123..9)&(ii for 10111213..99)and got 900 as result)))

=(300 + 900)

........
........
........

900901902....999=(9x100 +(sum of digits 0102030405...99(i.e same as above we did in (i for 123..9)&(ii for 10111213..99)and got 900 as result)))

=(900 + 900)

Therefore sum of all digits in 100101102....999=(100+200+300+400+500+600+700+800+900+ 900x9)

=4500+8100

=12600



PRATYUSH ANAND
 ()

ii)If you analyse carefully in 101112...99

from 10 to 19 , 1 in tense place is used 10 times and 1to9 in once place is appeared (i.e 1112)13....19) along with

1(in tense place).

and this is same for

202122...29

303132....39

.......

......

909192....99

Thus,

sum 101112..19=(1x10+(1+2+3..9))

=10+45

sum 202122..29=(2x10+(1+2+3..9))

=20+45

.......

.....

sum 909192..99=(9x10+(1+2+3..9))

=90+45

Therefore sum of all digits in 101112...99=(10+20+30+40+50+60+70+80+90+ 45x9)

=450+405

=855



PRATYUSH ANAND
 ()

Hi 2 all,
Here we have to add all the digits of a number P =1234567891011.............9989991000 as per question-->

i.e..

1,2,3,4,5,6,7,8,9,1,0,1,1,......9,9,9,1,0,0,0.

Now we will apply the formula of A.P

$S_n=\dfrac{n}{2}(2a+(n-1)d)$

We will divide the number in 4 parts -->

i) 1234..9

ii)101112...99

iii)100101102....999

iv)1000

i) Sum of number all digits from 123...9=9/2(2x1(9-1)1)=45



Mandar Dharmadhikari
 ()

can any1 tell me why the digit 9n unit place is 3????