Question

Find the least possible 3 digit number which when successively divided by 2, 5, 4 and 3 leaves the remainders 1, 1, 3 and 1 respectively.

• A. 273
• B. 197
• C. 193
• D. 527

Hint:

For successive division problems, always work backward from the last division. Choose the smallest non-negative final quotient (usually 0 or 1) that will eventually lead to a number in the desired range (e.g., 3-digit number).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the smallest 3-digit number that, when subjected to a sequence of successive divisions by 2, 5, 4, and 3, yields specific remainders (1, 1, 3, and 1, respectively). This is a number theory problem involving successive division.

Step 2: Key Formula or Approach:
Work backward from the last division.
If a number \( N \) is successively divided by \( d_1, d_2, d_3, \dots \) leaving remainders \( r_1, r_2, r_3, \dots \) and the final quotient is \( q_k \), then:
- The number before the last division is \( d_k \times q_k + r_k \).
- Continue working backward.
To find the least possible number, start with the smallest possible non-negative final quotient (usually 0 or 1).

Step 3: Detailed Explanation:
Let the number be N.
Divisors: 2, 5, 4, 3.
Remainders: 1, 1, 3, 1.
Let's denote the quotients as \( q_1, q_2, q_3, q_4 \).
$N \div 2 \rightarrow q_1$, remainder 1. So $N = 2q_1 + 1$.
$q_1 \div 5 \rightarrow q_2$, remainder 1. So $q_1 = 5q_2 + 1$.
$q_2 \div 4 \rightarrow q_3$, remainder 3. So $q_2 = 4q_3 + 3$.
$q_3 \div 3 \rightarrow q_4$, remainder 1. So $q_3 = 3q_4 + 1$.
To find the least possible N, we start with the least possible non-negative value for the final quotient, \( q_4 \).
Let \( q_4 = 1 \) (If $q_4=0$, $q_3=1$, $q_2=7$, $q_1=36$, $N=73$, which is not a 3-digit number).
If \( q_4 = 1 \):
$q_3 = 3(1) + 1 = 4$.
$q_2 = 4(4) + 3 = 16 + 3 = 19$.
$q_1 = 5(19) + 1 = 95 + 1 = 96$.
$N = 2(96) + 1 = 192 + 1 = 193$.
This number 193 is a 3-digit number. We can verify it:
$193 \div 2 = 96$ remainder 1.
$96 \div 5 = 19$ remainder 1.
$19 \div 4 = 4$ remainder 3.
$4 \div 3 = 1$ remainder 1.
All conditions are satisfied.
Since we started with the smallest possible \( q_4 \) (that yields a 3-digit number), 193 is the least possible 3-digit number.

Step 4: Final Answer:
The least possible 3-digit number is 193.

Answer 2

Since the question already gives four candidate numbers, a direct way to check each one is to actually carry out the successive divisions by 2, 5, 4, and 3, and see which candidate produces the remainders 1, 1, 3, and 1 in that exact order.

1. 273: Dividing by 2 gives quotient 136, remainder 1. Dividing 136 by 5 gives quotient 27, remainder 1. Dividing 27 by 4 gives quotient 6, remainder 3. Dividing 6 by 3 gives quotient 2, remainder 0, not the required remainder of 1, so 273 fails at the last division.
2. 197: Dividing by 2 gives quotient 98, remainder 1. Dividing 98 by 5 gives quotient 19, remainder 3, not the required remainder of 1, so 197 already fails at the second division.
3. 193: Dividing by 2 gives quotient 96, remainder 1. Dividing 96 by 5 gives quotient 19, remainder 1. Dividing 19 by 4 gives quotient 4, remainder 3. Dividing 4 by 3 gives quotient 1, remainder 1. All four remainders, 1, 1, 3, 1, match exactly, so 193 satisfies every condition.
4. 527: Dividing by 2 gives quotient 263, remainder 1, which matches the first condition. But dividing 263 by 5 gives quotient 52, remainder 3, not the required remainder of 1, so 527 does not satisfy the full chain either.

Since 193 is the only candidate that satisfies every successive remainder in the given order, it is the number that fits.

Therefore, the correct answer is 193.