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Last updated date: 28th Nov 2023
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# Find out the smallest number which when divided by any number from 2 to 10 the remainder will be 1.A. 2521B. 10000C. 5054D. 5500

Answer
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Hint: Type of question is based on the ‘lowest common factor’ which is abbreviated as LCM. According to the question we had to find out the smallest number which when divided by any of the numbers from 2 to 10(i.e. 2, 3, 4, 5, 6, 7, 8, 9, 10) then it will leave the remainder 1. In other words we have to find out the number which is divisible by all the numbers from 2 to 10 and always leave the remainder 1 when divided by these 2 to 10 numbers.

Complete step by step answer:
As concept says we will simply find the LCM of these numbers and add the value which we want as a remainder then we will get the lowest number which when divided by some given numbers then we will get the required remainder. So according to the given condition we will first find the LCM of numbers from 2 to 10 and add 1 to it as we want it as a remainder. So we will get the answer.
So we will get the answer by concept, we will find LCM of 2, 3, 4, 5, 6, 7, 8, 9, 10.
Factors of numbers are;
\begin{align} & 2=2 \\ & 3=3 \\ & 4=2\times 2 \\ & 5=5 \\ & 6=2\times 3 \\ & 7=7 \\ & 8=2\times 2\times 2 \\ & 9=3\times 3 \\ & 10=2\times 5 \\ \end{align}
So LCM of these numbers is;
$2\times 2\times 2\times 3\times 3\times 5\times 7=2520$
So 2520 is the LCM of numbers from 2 to 10. Now to get the value which always leaves a remainder 1 after division, so add 1 in 2520 so get the final answer. So we will get
\begin{align} & =2520+1 \\ & =2521 \\ \end{align}

So, the correct answer is “Option A”.

Note: While solving the value we get on finding the LCM of numbers from 2 to 10 i.e. 2520 will give the remainder zero when divided by any of the numbers ranging from 2 to 10.