Find the second term in a sequence of numbers that leaves the remainders 1, 2 and 7 when divided by 2, 3 and 8 respectively.

Correct option is C

Given:

A number leaves:

Remainder 1 when divided by 2

Remainder 2 when divided by 3

Remainder 7 when divided by 8
We are to find the second term in the sequence of such numbers.

Concept Used:

When the differences between divisors and their respective remainders are equal, i.e.,

2 - 1 = 3 - 2 = 8 - 7 = 1

this common difference dd can be used in the formula:

N = LCM(a,b,c) ⋅ k − d

Solution:

N = LCM(2, 3 , 8) ⋅ k − 1

LCM(2, 3, 8) = 24

N = 24k - 1

First term: k = 1 => N = 24⋅1 − 1 = 23

Second term: k = 2 => N = 24 ⋅ 2 − 1 = 47

​Alternate Solution:

Let the number be ( x ).

According to the problem, the number ( x ) satisfies the following congruences:

$$x \equiv 1 \pmod{2}$$​​

$$x \equiv 2 \pmod{3}$$​​

​$$x \equiv 7 \pmod{8}$$​​

To find the smallest positive solution for  x , we can solve the system of congruences step-by-step.

Solve  $$x \equiv 7 \pmod{8}$$​​

This implies that  x = 8k + 7  for some integer ( k ).

 Substitute ( x = 8k + 7 ) into the second congruence $$x \equiv 2 \pmod{3}$$​​

We get: $$8k + 7 \equiv 2 \pmod{3}$$​​

Simplify the expression modulo 3:

​$$8 \equiv 2 \pmod{3} \ so \ 8k + 7 \equiv 2k + 1 \pmod{3} \\ \\ 2k + 1 \equiv 2 \pmod{3}$$​​

Subtract 1 from both sides: $$2k \equiv 1 \pmod{3}$$​​

Multiply both sides by the modular inverse of 2 modulo 3, which is 2:  $$k \equiv 2 \pmod{3}$$​​

Thus, $$k = 3m + 2$$​for some integer ( m ).

Substitute ( k = 3m + 2 ) into ( x = 8k + 7 ):

 x = 8(3m + 2) + 7 

 x = 24m + 16 + 7 

 x = 24m + 23 

Therefore, the general solution for ( x ) is ( x = 24m + 23 ), where ( m ) is an integer.

 Find the second term by setting ( m = 1 ):

( x = 24(1) + 23 = 47 )

Answer: The second term in the sequence is ( 47 ).