Question

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

Answer 1

Hint: If we find the least common multiple of the numbers, we get the least possible number with remainder 0. So, now we find (Least common multiple + 1) as this leaves the remainder 1.

Complete step-by-step answer:
Now we will find Least Common Multiple of the following numbers: 2, 3, 4, 5 and 6.
Firstly, we do prime factorization of the following numbers 2, 3, 4, 5 and 6
Case 1: Prime factorization of 2
As 2 is itself a prime number we write 2 = 2
Case 2: Prime factorization of 3
As 3 is also itself a prime number we write 3 = 3
Case 3: Prime factorization of 4
As 4 can be written in terms of prime number 2
4 = 2.2
$4=2^2$
Case 4: Prime factorization of 5
As 5 is itself a prime number we write 5 = 5
Case 5: Prime factorization of 6
As 6 can be written in terms of 2 and 3 we write:
6 = 2.3
Now we will write all numbers with their prime factorized form, together:
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
So, we will remove the terms which are repeated. Now we will write the repeated terms: One of the 2 is repeated thrice (in 2, 4, 6). The number 3 is repeated twice (in 3, 6). By removing the repeated terms and writing them only once we get:
Least Common Multiple = 2. 3. 2. 5
By simplifying we get:
Least common multiple = 6. 10
Least common multiple = 60
As 60 is divisible by the numbers 2, 3, 4, 5, 6 we get:
a 0 remainder when 60 is divided by numbers 2, 3, 4, 5, 6.
So, if we add 1 to the Least Common Multiple, we get:
Least Common Multiple + 1 = 61
61 when divided by 2, 3, 4, 5, 6 gives remainder 1.
but the second question was the number must be divisible by 7.
So, now we need to multiply an integer to least common multiple and then add 1, to make both conditions satisfy.
$$Requirednumber=[\left(LeastCommonMultiple\right).n]+1$$
Case 1: n = 2
By substituting n value in above equation, we get:
Required number = (60 x 2) + 1
             = 121
121 is not divisible by 7.
Case 2: n = 3
By substituting n value in above equation, we get:
Required number = (60 x 3) + 1
             = 181
181 is not divisible by 7.
Case 3: n = 4
By substituting the value of n in above equation, we get:
Required number = (60 x 4) + 1
             = 241
241 is not divisible by 7.
Case 4: n = 5
By substituting the value of n in above equation, we get:
Required number = (60 x 5) + 1
             = 301
301 is divisible by 7.
Therefore 301 is the least number which leaves remainder as 1 when divided by 2, 3, 4, 5, 6 and leaves no remainder when divided by 7.
Option (b) is correct.

Note: While taking Least Common Multiple be careful to remove only repeated numbers.
Getting an idea to add 1 to the Least common multiple is crucial. So, be careful while applying it.