Problem 21 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The digit at unit's place in the number \(17^{1995}+11^{1995}\) \(-7^{1995}\) is (A) 0 (B) 1 (C) 2 (D) 3

Step-by-Step Solution

Answer

The unit's digit of the expression is 1.

Step 1: Understand the Problem

We need to find the digit at the unit's place of the expression \(17^{1995} + 11^{1995} - 7^{1995}\). This involves computing the unit's place of each power term separately and then combining them.

Step 2: Determine Unit's Digit Cycle

First, identify the cycle patterns for the units digits in terms of powers: - **For 17:** The units place cycle for powers of 7 repeats every 4: 7, 9, 3, 1. - **For 11:** The units digit of any power of 11 is always 1, since 1 raised to any power is 1. - **For 7:** The unit's digit cycle for powers of 7 also repeats every 4: 7, 9, 3, 1.

Step 3: Compute Remainder for Exponents

Since the cycles repeat every 4, divide the exponent 1995 by 4 to find the position within the cycle: - The remainder when 1995 is divided by 4 is 3, because 1995 divided by 4 gives a quotient of 498 with a remainder of 3.

Step 4: Find Unit's Digits Using Remainder

Using the remainder of 3:- For \(17^{1995}\), the third term of the cycle is 3.- For \(11^{1995}\), the unit's digit is always 1.- For \(7^{1995}\), the third term of the cycle is 3.

Step 5: Apply Expression to Unit's Digits

Substitute and simplify the expression using the calculated unit's digits:\[3 (from \ 17^{1995}) + 1 (from \ 11^{1995}) - 3 (from \ 7^{1995}) = 1\]

Step 6: Conclusion

The unit digit of \(17^{1995} + 11^{1995} - 7^{1995}\) is 1.

Key Concepts

Exponentiation

When dealing with large numbers and expressions like \(17^{1995} + 11^{1995} - 7^{1995}\), it's crucial to understand the concept of exponentiation. Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The base is raised to the power of the exponent, which tells you how many times to multiply the base by itself. In simple terms, for \(a^n\), you multiply \(a\) by itself \(n\) times. For example:

• \(2^3 = 2 \times 2 \times 2 = 8\)
• \(5^4 = 5 \times 5 \times 5 \times 5 = 625\)

In the original exercise, we face very large exponents like \(1995\). Instead of calculating the entire power directly, which would be computationally heavy and inefficient, we use patterns in the digits to simplify the process.

Remainder Calculation

Understanding remainder calculation is key in determining the position of a digit in its cycle pattern. When we divide a number, the remainder is what is left after the division.For instance, if you divide 10 by 3, the quotient is 3, and the remainder is 1. This is written as:

• \(10 \div 3 = 3\) remainder \(1\)

In the context of the problem, remainder calculation helps us to find our position within a repetitive cycle by dividing the exponent by the cycle length. Here, the cycle length is 4.By calculating the remainder of \(1995 \div 4\), we determine that the remainder is 3. This means that for each base, we should look at the third position in its respective cycle pattern to find the unit's digit.

Cycle Pattern

In mathematics, especially in number theory, recognizing cycle patterns is a valuable strategy for simplifying calculations. A cycle pattern refers to the repeating sequence of numbers that appear as the unit's digit when powers of a number are considered. For example, let's look at a few base numbers:

• **For base 7**: The pattern in its powers ends in the digits 7, 9, 3, 1 which repeats every 4 terms.
• **For base 11**: Powers of 11 always end with the digit 1 because any power of 1 (unit place of 11) remains 1.
• **For base 17**: It follows the same pattern as base 7, with a cycle of 7, 9, 3, 1.

To solve the problem, these cycle patterns allow us to determine the correct unit digit without extensive calculation by observing the remainder from division with cycle length. Knowing that the remainder was 3 from previous calculation, we simply took the third digit from each cycle pattern to easily find the unit's digits: for \(17^{1995}\) and \(7^{1995}\), both end in 3, and for \(11^{1995}\), it is always 1. This simplification greatly aids in quickly evaluating such problems without complex computation.