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

Step 1: Understand successive division.
When a number is divided successively, the quotient from one step becomes the new number for the next step. Here we divide by $2$, then $3$, then $5$, getting remainders $1$, $2$, $3$ in that order. We want the smallest such number that also fits the extra clue.

Step 2: Build the number from the last step.
Work backwards. Start with the smallest quotient we can pick. The last division by $5$ gives remainder $3$, so the value entering that step is $5q + 3$. Take the simplest case $q = 2$, giving $5 \times 2 + 3 = 13$.

Step 3: Go back through the divide by $3$ step.
That $13$ was the quotient of the second division, which had remainder $2$. So the value entering the second step is $3 \times 13 + 2 = 41$.

Step 4: Go back through the divide by $2$ step.
That $41$ was the quotient of the first division, which had remainder $1$. So the original number is $2 \times 41 + 1 = 83$.

Step 5: Check the remainder pattern.
Divide $83$ by $2$: quotient $41$, remainder $1$. Divide $41$ by $3$: quotient $13$, remainder $2$. Divide $13$ by $5$: quotient $2$, remainder $3$. All three remainders match.

Step 6: Confirm it is under $200$.
The value $83$ is well below $200$ and is the number this successive pattern produces. So the required number is \[ \boxed{83} \]