Find the number when successively divided by 3,5, and 7 leaves the remainder 2,1, and 3, respectively, and the last quotient is 3.

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SSC CHSL Exam 2024 Tier-I Official Paper (Held On: 01 Jul, 2024 Shift 2)
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  1. 367
  2. 360
  3. 365
  4. 362

Answer (Detailed Solution Below)

Option 3 : 365
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Successive division = 3, 5, and 7

Successive remainder = 2, 1, and 3

Let the last quotient = x

then number will be 

3 [5 {7x + 3} + 1] + 2

⇒ 3 [35x + 15 + 1] + 2

⇒ 105x + 45 + 3 + 2

⇒ 105x + 50

⇒ 5[21x + 10]

Since, the obtained number is a multiple of 5 then according to the option either 360 or 365 will be the correct answer.

Now, if number = 365

105x + 50 = 365

⇒ 105x = 315

⇒ x = 3

Here, we get the integer value of x, so the correct number will be 365.

∴ The correct answer is option (3).

Alternate Method

Given:

Number when successively divided by 3, 5, and 7 leaves remainders 2, 1, and 3, respectively, and the last quotient is 3.

Formula Used:

Chinese Remainder Theorem

Calculation:

Let's denote the number as ( N ).

According to the problem, when ( N ) is successively divided by 3, 5, and 7, it leaves remainders 2, 1, and 3, respectively, with the last quotient being 3.

We can break this down into a step-by-step process.

Let’s consider the successive division:

First Division by 3: ( N ) when divided by 3 leaves a remainder of 2. Thus, N = 3k + 2 for some integer k.

Second Division by 5: The quotient ( k ), obtained from the first division, when divided by 5, leaves a remainder of 1. Thus, k = 5m + 1 for some integer m.

Substituting back into the equation for ( N ): N = 3(5m + 1) + 2 = 15m + 3 + 2 = 15m + 5.

Third Division by 7: The quotient ( m ), obtained from the second division, when divided by 7, leaves a remainder of 3. Thus, m = 7n + 3 for some integer n.

Substituting back into the equation for ( N ): N = 15(7n + 3) + 5 = 105n + 45 + 5 = 105n + 50.

Last Quotient: It’s given that the last quotient from the third division is 3.

Hence, n = 3. Substituting (n = 3) into the equation for N : N = 105(3) + 50 = 315 + 50 = 365.

Thus, the number ( N ) that satisfies all the conditions is 365.

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