Question

Find the number nearest to “100000” and greater than “100000” which is exactly divisible by each of “8”,”15” and “21”?
A.100800
B.100600
C.110800
D.101800

Answer 1

Hint: To get the solution of such a question you need to find the lowest common factor of each all the given divisors, and once you get it then divide the given numbers with the obtained number to get the result. Lowest common factors are the values for the given numbers which can satisfy the condition of complete division of a number by the given set of numbers.

Complete step-by-step answer:
The number which has to be find should be greater than “100000”
Firstly finding the lowest common factor of the divisor “8”,”15”,”21” we get:
\[
   \Rightarrow factors\,of\,8 = 1,2,4 \\
   \Rightarrow factors\,of\,15 = 1,3,5 \\
   \Rightarrow factors\,of\,21 = 1,3,7 \\
   \Rightarrow lowest\,common\,factor\,of\,these\,numbers\,are = 1 \times 2 \times 4 \times 3 \times 5 \times 7 = 840 \;
 \]
Here we found the lowest common factor, now on dividing each term we have to check for exact divisible, on solving we get:
\[
   \Rightarrow the\,first\,given\,number\,is\,100800 \\
   \Rightarrow on\,dividing\,with\,L.C.Mwe\,get\, \\
   \Rightarrow \dfrac{{100800}}{{840}} = 120\,which\,is\,exact\,divisible \;
 \]
Now when we check for other numbers, we do not get the exact divisible, so “100800” is our right answer.
So, the correct answer is “Option A”.

Note: The given number is very large to think for, but for small numbers you can directly check the divisibility by dividing the numbers separately by each term and get the solution.
For finding the lowest common factor you have to factorize the numbers and then get the common factors as individual factors, the common factors obtained will be count for one time and multiplied only one time with the rest of the factors obtained, the final product is our solution.