The least number which when divided by 2,3,4,5, and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder. The number is
A) 231
B) 301
C) 371
D) 441

Hint: In the given question, we want to find the smallest number divided by 2,3,4,5, and 6 leaves remainder 1 and completely divide by 7. Assume the required number with variable x. The second step is to find the least common multiple of numbers 2, 3, 4, 5, and 6. The number is divided by 7. So, by substituting different values we can find the required number.

Complete step-by-step solution:
Here we want to kind the unknown number.
Therefore, let us assume that the number will be x.
When the number is divided by 2,3,4,5, and 6 leaves the remainder 1 in each case.
$ \Rightarrow x - 1$
Therefore, we find the least common multiple of 2,3,4,5, and 6.
$LCM\left( {2,3,4,5,6} \right) = 60$
So, we can write.
$ \Rightarrow x - 1 = 60n$
Where the value of n is a natural number 1, 2, 3, ...
Add 1 on both sides.
$ \Rightarrow x = 60n + 1$
Here, the number is divided by 7.
$ \Rightarrow x = \left( {\dfrac{{60n + 1}}{7}} \right)$
Now, substitute different values of n and find x.
 First, take the value of n is 1.
$ \Rightarrow x = \left( {\dfrac{{60\left( 1 \right) + 1}}{7}} \right)$
$ \Rightarrow x = \dfrac{{61}}{7}$
This is not divided by 7.
Now, take the value of n is 3.
$ \Rightarrow x = \left( {\dfrac{{60\left( 2 \right) + 1}}{7}} \right)$
$ \Rightarrow x = \dfrac{{121}}{7}$
This is not divided by 7.
Now, take the value of n is 3.
$ \Rightarrow x = \left( {\dfrac{{60\left( 3 \right) + 1}}{7}} \right)$
$ \Rightarrow x = \dfrac{{181}}{7}$
This is not divided by 7.
Now, take the value of n is 4.
$ \Rightarrow x = \left( {\dfrac{{60\left( 4 \right) + 1}}{7}} \right)$
$ \Rightarrow x = \dfrac{{241}}{7}$
This is not divided by 7.
Now, take the value of n is 5.
$ \Rightarrow x = \left( {\dfrac{{60\left( 5 \right) + 1}}{7}} \right)$
$ \Rightarrow x = \dfrac{{301}}{7}$
This is divided by 7.
Hence, the required number is 301. When we divide the number 301 with 2,3,4,5, and 6 leaves remainder 1, and completely divide by 7.

Option B is the correct answer.

Note: Here, make sure that the equation which we construct must be correct. And find the least common multiple of given positive numbers which is divisible completely by all given numbers. The LCM of two or more numbers is greater than or equal to the greatest number of given numbers.