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Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 95)
Numbers
95.
The smallest 5 digit number exactly divisible by 41 is:
1004
10004
10045
10025
None of these
Answer: Option
Explanation:
The smallest 5-digit number = 10000.

 41) 10000 (243
     82
     ---
     180
     164
     ----
      160
      123
      ---
       37
      --- 

 Required number = 10000 + (41 - 37)
                 = 10004.
Discussion:
6 comments Page 1 of 1.
Surya Teja said:   7 years ago
On dividing the 10000 with 41 we are getting remainder of 37, so we have to subtract the 37 from 10000 to get perfectly divisible by 41 but it will be 4 digit number so that again we have to add 41 for smallest 5 digit number divisible by 41.
(1)
Rita said:   1 decade ago
Why we need to subtract from (41-37)?
Naseer said:   10 years ago
How we can subtract from 85?
Ani said:   8 years ago
Why we need to subtract from (41-37)?
Mahendi said:   7 years ago
How 82 came?
Sasi said:   2 years ago
How come 243? Please explain.
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