Latest Elitmus Aptitude Question SOLUTION: Z is a number in decimal system such that Z=260*1024+73*512+128*129+81+9. Let Y be the octal representation of 'Z' (that is representation of 'z' in a number system to the base 8 i

• Placement Papers
• /
• Elitmus
• /
• Logical Reasoning
• /
• Number Series

Elitmus Question Papers:

(#M40018184) ELITMUS Logical Reasoning question ELITMUS MATHS PUZZLE Keep an EYE

Z is a number in decimal system such that Z=260*1024+73*512+128*129+81+9. Let Y be the octal representation of 'Z' (that is representation of 'z' in a number system to the base 8 is 'Y'). How many times will the digit 3 be there in 'Y'

Asked In Elitmus pavan (11 years ago)

Unsolved Read Solution (4)

Is this Puzzle helpful?   (6)   (2) Submit Your Solution Number Series

Read Solution (4) :

taashi  (11 years ago)

260*1024+73*512+129*128+90->2
260*128+73*64+129*16+11->3
260*16+73*8+259->32
625->1
78->6
9->1
1->1

reading from down to top 1161332.
so there are two 3's.

Is this solution Helpfull? Yes (24) | No (4)  

GANESH PRASAD  (8 years ago)

@Taashi how 2,3,32 are forming please explain

Is this solution Helpfull? Yes | No   

ramu nagisetty  (8 years ago)

@ganesh when we dive by 8 the remaining sum is 2,3,3 its not 32 dont confuse

Is this solution Helpfull? Yes | No   

TEJA VARDHAN REDDY  (7 years ago)

Given,
Z=260*1024+73*512+128*129+81+9(which is in decimal form)

Convert each number in to octal form by placing remainders......
Let's take for (260)base to the 10 = (404)base to the 8
Similarily 1024= 2000
......

Finally octal form of z is
Z=404*2000+111*1000+200*201+121+11

---111=3*37 && 201=3*67

Z=404*2000+(3*37)*1000+200*(3*67)+121+11

How many times the digit 3 is repeated.......
2 times

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.