Number puzzle: which 5-digit number becomes exactly three times larger when the digit 1 is written at the end instead of at the beginning?

Introduction / Context:
This classic number puzzle tests logical reasoning with place value and simple algebra. We are given a special 5 digit number such that writing the digit 1 in front of it produces a number that is exactly three times smaller than the number obtained by writing the digit 1 at the end. Questions like this often appear in aptitude tests and bank or SSC style exams to check comfort with numbers and proportional relationships.

Given Data / Assumptions:

• The original number is a 5 digit integer. Let us call it N.
• Writing 1 at the beginning gives a 6 digit number, written as 1N.
• Writing 1 at the end gives a 6 digit number, written as N1.
• The number with 1 at the end is three times the number with 1 at the beginning.
• We assume standard base 10 notation and no leading zeros in N.

Concept / Approach:
We convert the verbal description into equations using place value. If N is 5 digits, then putting 1 in front is 1 * 10^5 + N, and putting 1 at the end is 10 * N + 1. The condition that the second is three times the first becomes a linear equation in N. Solving this equation gives the required number. Finally, we verify that the solution is a 5 digit integer and that it indeed satisfies the given relationship exactly.

Step-by-Step Solution:
Let N be the unknown 5 digit number. Number with 1 at the front = 1 * 10^5 + N = 100000 + N. Number with 1 at the end = 10 * N + 1. Given condition: 10 * N + 1 = 3 * (100000 + N). Expand right side: 10 * N + 1 = 300000 + 3 * N. Rearrange: 10 * N - 3 * N = 300000 - 1. So 7 * N = 299999. Therefore N = 299999 / 7 = 42857. Check that N is 5 digits and satisfies the original wording.

Verification / Alternative check:
Compute the two 6 digit numbers explicitly. With N = 42857, putting 1 in front gives 142857. Putting 1 at the end gives 428571. Now check the ratio: 428571 / 142857 = 3 exactly, with no remainder. This shows that the number obtained by placing 1 at the end is exactly three times the number obtained by placing 1 at the beginning. Also, 42857 clearly has 5 digits, so it satisfies all required conditions. No other 5 digit integer will satisfy the linear equation 7 * N = 299999, so the solution is unique.

Why Other Options Are Wrong:
Option 14285 does not work: 114285 * 3 equals 342855, which is not equal to 142851. Option 28571 fails in the same way because the ratio between 128571 and 285711 is not exactly three. Option 42168 also does not satisfy the algebraic equation because it does not solve 7 * N = 299999. Only 42857 produces the precise three times relationship demanded in the problem statement.

Common Pitfalls:
A common error is to misinterpret three times smaller and set up the proportion incorrectly, for example by writing 1N = 3 * N1 instead of the correct 10 * N + 1 = 3 * (100000 + N). Another frequent mistake is mishandling powers of 10 and place values when constructing 1N and N1. Some students also attempt guess and check with random 5 digit numbers, which is inefficient and prone to arithmetic mistakes. Setting up the clear algebraic equation and solving systematically is the safest and fastest method in exam conditions.

Final Answer:
The required 5 digit number is 42857.