Latest Elitmus Aptitude Question SOLUTION: z is a number in decimal system that z = 260*1024 + 73*512 +128*129 +81 +9.Let y be te octal representation of z .how many times will the digit 3 be there in y . option 1. 2 2. 3 3

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(#M40030479) ELITMUS Numerical Ability question Keep an EYE

z is a number in decimal system that z = 260*1024 + 73*512 +128*129 +81 +9.Let y be te octal representation of z .how many times will the digit 3 be there in y .
option
1. 2
2. 3
3. 4
4. 5

Asked In Elitmus rahul kumar (10 years ago)

Solved Mahesh Read Solution (11)

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Read Solution (11) :

Ankit Goyal  (10 years ago)

The above decimal expression can be written as:-

8(260*128 + 73*64 + 129*16 + 11) + 2 -- take aside remainder 2
8(260*16 + 73*8 + 129*2 + 1) + 3 -- take aside remainders 32
8(260*2 + 73 + 32) + 3 -- remainders 332
8(65 + 4 + 9) + 1 -- remainders 1332
8(9) + 6 -- remainders 61332
8(1) + 1 - remainders 161332
finally the number ion octal = 1161332
Ans : 2

Is this solution Helpfull? Yes (45) | No   

kirti agarwal  (10 years ago)

easy solution for dis question is like:
since y is octal represention of z, and z is decimal so 1st solve equation z.
z=(260*1024)+(73*512)+(128*129)+81+9
= 26624+ 37276 + 16512 +81 + 9
= 80602
Now convert dis 80602 in octal by dividing dis number by 8,
u will get 234132

so 3 exist 2times.

Is this solution Helpfull? Yes (21) | No (11)  

Mahesh  (10 years ago)

First lets convert everything into the highest power of 8 visible in the question, that is 512.

S0, it can be written as 520 * 512 + 73 * 512 + 128 * (128+1), the reason for changing the 128 expression is to convert it also into 512.
So, the expression is 512 * 520 + 512 * 32 + 512 * 32( which is 128/4)

So, that will result as 512 * 625
Now this will help in finding all the digits from 4 th digit, the method is
dividing 625 by 8, the quotient is 78 and remainder is 1
dividing 78 by 8, the quotient is 9 and the remainder is 6
dividing 9 by 8, the quotient is 1 and the remainder is 1

So, the first four (from right) digits of the octal expression are 1161.

Now the sum of remaining three numbers is 128 + 81 +9 = 218

If we divide it by 8, the quotient is 27 and remainder 2
If 27 is divided by 8, the quotient is 3 and remainder is 3

So, the last three digits are 332

So, the entire number is 1161332
The number of 3's is two
Answer is two

Is this solution Helpfull? Yes (9) | No   

shalini  (10 years ago)

260*(2^10)+ 73*(2^9)+129*(2^7)+90
2^7*(260*8+73*4+129)+90
16*8*(2501)+90
16*20008+90
320128+90=320218
change this to octal value by deviding 8 you will get (1161332)base8
here we got 2 times 3.

Is this solution Helpfull? Yes (5) | No   

shobhit agarwal  (10 years ago)

solving the value for z we get
z= 80602
converting it to octal we get 235332

So the answer is 3 digit will be there 3 times.

Is this solution Helpfull? Yes (2) | No (3)  

salini sharma  (10 years ago)

ARCHNA Explain How?

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archna  (10 years ago)

1. 2 3's
just convert z into binary and then convert binary into octal,and see the total no. of 3's

Is this solution Helpfull? Yes | No (5)  

Himanshu Soni  (10 years ago)

ise bina binary me convert kiye bhi direct octal me convert kiya ja sakta hai.

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Abhishek Kumar  (10 years ago)

Himanshu Kaise????

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pankaj kumar  (10 years ago)

i guess its 4

Is this solution Helpfull? Yes | No (1)  

rajveer singh  (8 years ago)

Normally multiplying numbers in z, we get 316291
and then converting it to octal we get 111132 which means only one 3 is there.
so options are wrong.

Is this solution Helpfull? Yes | No (1)  

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