Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 95)
95.
The smallest 5 digit number exactly divisible by 41 is:
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:
5 comments Page 1 of 1.
Surya Teja said:
4 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.
Mahendi said:
5 years ago
How 82 came?
Ani said:
5 years ago
Why we need to subtract from (41-37)?
Naseer said:
7 years ago
How we can subtract from 85?
Rita said:
8 years ago
Why we need to subtract from (41-37)?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers