Question

A number when divided successfully by 2, 3, and 5 leaves remainders 1, 2, and 3 respectively. If the number is less than 200 and also divisible by 7 after adding 4 to it, then that number is

• A. (53)
• B. (83)
• C. (143)
• D. (173)

Hint:

To easily find the general expression for a number undergoing successive divisions, use the bottom-up synthetic method: \[ \text{Number} = (((k \times d_3 + r_3) \times d_2 + r_2) \times d_1 + r_1) \] Always test options directly with the final given conditions (like adding 4 and checking divisibility by 7) to save precious time during competitive exams!

Answer 1

The Correct Option is B

Concept: The problem involves successive division (also called consecutive division). When a number (N) is divided successfully by divisors (d_1, d_2, d_3) leaving remainders (r_1, r_2, r_3), it means:

• (N) divided by (d_1) gives a quotient (q_1) and remainder (r_1).

• (q_1) divided by (d_2) gives a quotient (q_2) and remainder (r_2).

• (q_2) divided by (d_3) gives a final quotient (q_3) and remainder (r_3).
We can reconstruct the general expression for the number (N) by working backwards from the last quotient.

Step 1: Reconstruct the number from the successive division steps. Let the final quotient obtained after division by the last divisor (5) be (k), where (k 0) is an integer.

• According to the last division step by 5 with remainder 3:

• According to the second division step by 3 with remainder 2:
Substituting the value of (q_2) into this equation:

• According to the first division step by 2 with remainder 1: \[ N = 2q_1 + 1 \] Substituting the value of (q_1) into this equation: \[ N = 2(15k + 11) + 1 = 30k + 22 + 1 = 30k + 23 \] Thus, the general form of the number is (N = 30k + 23), where (k 0, 1, 2, 3, ).

Step 2: Find the possible values of (N) that are less than 200. Let us substitute different non-negative integers for (k): itemize

• For (k = 0): (N = 30(0) + 23 = 23)

• For (k = 1): (N = 30(1) + 23 = 53)

• For (k = 2): (N = 30(2) + 23 = 83)

• For (k = 3): (N = 30(3) + 23 = 113)

• For (k = 4): (N = 30(4) + 23 = 143)

• For (k = 5): (N = 30(5) + 23 = 173)

• For (k = 6): (N = 30(6) + 23 = 203) (This is greater than 200, so we stop here).
The possible numbers less than 200 are: (23, 53, 83, 113, 143, 173).

Step 3: Apply the final condition: the number becomes divisible by 7 after adding 4 to it. This means ((N + 4)) must be a multiple of 7. Let's test our candidates:

• For (N = 23): (23 + 4 = 27), which is not divisible by 7.

• For (N = 53): (53 + 4 = 57), which is not divisible by 7.

• For (N = 83): (83 + 4 = 87), which is not divisible by 7. Wait, let's re-verify the division: (87 / 7 = 12.42) (not divisible).

• Let's re-read carefully: (N = 83 83 + 4 = 87). Let's check all numbers systematically: itemize

• (23 + 4 = 27) (No)

• (53 + 4 = 57) (No)

• (83 + 4 = 87) (No)

• (113 + 4 = 117) (No)

• (143 + 4 = 147). Let's check $(147 \div 7 = 21)$. Yes! (147) is perfectly divisible by 7.

• (173 + 4 = 177) (No)
itemize Let's double check the calculation to ensure option matching or any translation discrepancy. The question text states: "If the number is less than 200 and also divisible by 7 after adding 4 to it, then that number is". Testing (N = 143): (143 + 4 = 147), which is (7 21). Therefore, (143) perfectly satisfies all given criteria.