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

Each candidate number can be checked directly by performing the successive divisions by 2, 5, 4, and 3 in order and comparing the remainders obtained at each stage against the required sequence 1, 1, 3, 1.

1. 273: \( 273 \div 2 = 136 \) remainder \( 1 \). \( 136 \div 5 = 27 \) remainder \( 1 \). \( 27 \div 4 = 6 \) remainder \( 3 \). \( 6 \div 3 = 2 \) remainder \( 0 \), not the required remainder of 1, so 273 fails at the last step.
2. 197: \( 197 \div 2 = 98 \) remainder \( 1 \). \( 98 \div 5 = 19 \) remainder \( 3 \), not the required remainder of 1, so 197 already fails at the second step.
3. 193: \( 193 \div 2 = 96 \) remainder \( 1 \). \( 96 \div 5 = 19 \) remainder \( 1 \). \( 19 \div 4 = 4 \) remainder \( 3 \). \( 4 \div 3 = 1 \) remainder \( 1 \). Every step matches the required remainders of 1, 1, 3, 1 exactly.
4. 527: \( 527 \div 2 = 263 \) remainder \( 1 \). \( 263 \div 5 = 52 \) remainder \( 3 \), not the required remainder of 1, so 527 fails at the second step, and is also a larger 3-digit number than 193 in any case.

Only 193 produces exactly the remainder sequence 1, 1, 3, 1 through all four successive divisions, and it is the smallest such 3-digit number among the options.

Therefore, the correct answer is 193.