Understanding the Problem
We are asked to find the number of different possible remainders when the sum \(1^a + 2^b + 3^c + 4^d\) is divided by 10, where \(a, b, c, d\) are natural numbers. Natural numbers are positive integers starting from 1 (i.e., \(1, 2, 3, \dots\)). Finding the remainder when dividing by 10 is the same as finding the last digit of the number.
Analyzing the Powers and Their Last Digits (Remainders)
The last digit of powers of a number follows a pattern, known as cyclicity. Let's find the possible last digits for each term in the sum:
- Last digit of \(1^a\): For any natural number \(a\), \(1^a\) is always 1. So, the remainder is always 1. \(1^a \pmod{10} \equiv 1\)
- Last digit of \(2^b\): The last digits of powers of 2 follow a cycle: \(2^1 = 2\) \(2^2 = 4\) \(2^3 = 8\) \(2^4 = 16 \implies 6\) \(2^5 = 32 \implies 2\) The cycle of the last digits is (2, 4, 8, 6). This cycle has a length of 4. Since \(b\) can be any natural number, the possible remainders for \(2^b\) when divided by 10 are \(\{2, 4, 8, 6\}\).
- Last digit of \(3^c\): The last digits of powers of 3 follow a cycle: \(3^1 = 3\) \(3^2 = 9\) \(3^3 = 27 \implies 7\) \(3^4 = 81 \implies 1\) \(3^5 = 243 \implies 3\) The cycle of the last digits is (3, 9, 7, 1). This cycle also has a length of 4. The possible remainders for \(3^c\) when divided by 10 are \(\{3, 9, 7, 1\}\).
- Last digit of \(4^d\): The last digits of powers of 4 follow a cycle: \(4^1 = 4\) \(4^2 = 16 \implies 6\) \(4^3 = 64 \implies 4\) The cycle of the last digits is (4, 6). This cycle has a length of 2. The possible remainders for \(4^d\) when divided by 10 are \(\{4, 6\}\).
Finding Possible Remainders of the Sum
We need to find the possible remainders of the sum \(S = 1^a + 2^b + 3^c + 4^d\) when divided by 10. This is equivalent to finding the possible values of \((1^a \pmod{10} + 2^b \pmod{10} + 3^c \pmod{10} + 4^d \pmod{10}) \pmod{10}\).
Let \(R_1, R_2, R_3, R_4\) be the remainders of \(1^a, 2^b, 3^c, 4^d\) respectively when divided by 10.
- \(R_1 = 1\)
- \(R_2 \in \{2, 4, 8, 6\}\)
- \(R_3 \in \{3, 9, 7, 1\}\)
- \(R_4 \in \{4, 6\}\)
We need to find the possible values of \((R_1 + R_2 + R_3 + R_4) \pmod{10}\). Let's analyze the possible sums by considering the possible values for \(R_4\). Since \(R_1\) is always 1, the expression becomes \((1 + R_2 + R_3 + R_4) \pmod{10}\).
Case 1: \(R_4 = 4\)
The sum's remainder modulo 10 is \((1 + R_2 + R_3 + 4) \pmod{10} = (5 + R_2 + R_3) \pmod{10}\).
- If \(R_2 = 2\):
- \(R_3 = 3 \implies (5 + 2 + 3) \pmod{10} = 10 \pmod{10} = 0\)
- \(R_3 = 9 \implies (5 + 2 + 9) \pmod{10} = 16 \pmod{10} = 6\)
- \(R_3 = 7 \implies (5 + 2 + 7) \pmod{10} = 14 \pmod{10} = 4\)
- \(R_3 = 1 \implies (5 + 2 + 1) \pmod{10} = 8 \pmod{10} = 8\)
- If \(R_2 = 4\):
- \(R_3 = 3 \implies (5 + 4 + 3) \pmod{10} = 12 \pmod{10} = 2\)
- \(R_3 = 9 \implies (5 + 4 + 9) \pmod{10} = 18 \pmod{10} = 8\)
- \(R_3 = 7 \implies (5 + 4 + 7) \pmod{10} = 16 \pmod{10} = 6\)
- \(R_3 = 1 \implies (5 + 4 + 1) \pmod{10} = 10 \pmod{10} = 0\)
- If \(R_2 = 8\):
- \(R_3 = 3 \implies (5 + 8 + 3) \pmod{10} = 16 \pmod{10} = 6\)
- \(R_3 = 9 \implies (5 + 8 + 9) \pmod{10} = 22 \pmod{10} = 2\)
- \(R_3 = 7 \implies (5 + 8 + 7) \pmod{10} = 20 \pmod{10} = 0\)
- \(R_3 = 1 \implies (5 + 8 + 1) \pmod{10} = 14 \pmod{10} = 4\)
- If \(R_2 = 6\):
- \(R_3 = 3 \implies (5 + 6 + 3) \pmod{10} = 14 \pmod{10} = 4\)
- \(R_3 = 9 \implies (5 + 6 + 9) \pmod{10} = 20 \pmod{10} = 0\)
- \(R_3 = 7 \implies (5 + 6 + 7) \pmod{10} = 18 \pmod{10} = 8\)
- \(R_3 = 1 \implies (5 + 6 + 1) \pmod{10} = 12 \pmod{10} = 2\)
In this case (\(R_4 = 4\)), the possible remainders are \(\{0, 2, 4, 6, 8\}\).
Case 2: \(R_4 = 6\)
The sum's remainder modulo 10 is \((1 + R_2 + R_3 + 6) \pmod{10} = (7 + R_2 + R_3) \pmod{10}\).
- If \(R_2 = 2\):
- \(R_3 = 3 \implies (7 + 2 + 3) \pmod{10} = 12 \pmod{10} = 2\)
- \(R_3 = 9 \implies (7 + 2 + 9) \pmod{10} = 18 \pmod{10} = 8\)
- \(R_3 = 7 \implies (7 + 2 + 7) \pmod{10} = 16 \pmod{10} = 6\)
- \(R_3 = 1 \implies (7 + 2 + 1) \pmod{10} = 10 \pmod{10} = 0\)
- If \(R_2 = 4\):
- \(R_3 = 3 \implies (7 + 4 + 3) \pmod{10} = 14 \pmod{10} = 4\)
- \(R_3 = 9 \implies (7 + 4 + 9) \pmod{10} = 20 \pmod{10} = 0\)
- \(R_3 = 7 \implies (7 + 4 + 7) \pmod{10} = 18 \pmod{10} = 8\)
- \(R_3 = 1 \implies (7 + 4 + 1) \pmod{10} = 12 \pmod{10} = 2\)
- If \(R_2 = 8\):
- \(R_3 = 3 \implies (7 + 8 + 3) \pmod{10} = 18 \pmod{10} = 8\)
- \(R_3 = 9 \implies (7 + 8 + 9) \pmod{10} = 24 \pmod{10} = 4\)
- \(R_3 = 7 \implies (7 + 8 + 7) \pmod{10} = 22 \pmod{10} = 2\)
- \(R_3 = 1 \implies (7 + 8 + 1) \pmod{10} = 16 \pmod{10} = 6\)
- If \(R_2 = 6\):
- \(R_3 = 3 \implies (7 + 6 + 3) \pmod{10} = 16 \pmod{10} = 6\)
- \(R_3 = 9 \implies (7 + 6 + 9) \pmod{10} = 22 \pmod{10} = 2\)
- \(R_3 = 7 \implies (7 + 6 + 7) \pmod{10} = 20 \pmod{10} = 0\)
- \(R_3 = 1 \implies (7 + 6 + 1) \pmod{10} = 14 \pmod{10} = 4\)
In this case (\(R_4 = 6\)), the possible remainders are also \(\{0, 2, 4, 6, 8\}\).
Conclusion
Combining the results from both cases, the set of all possible remainders when \(1^a + 2^b + 3^c + 4^d\) is divided by 10 is \(\{0, 2, 4, 6, 8\}\).
There are 5 distinct possible remainders.
Similar Questions
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Important Questions from Divisibility and Remainder
If the 8-digit number 888x53y4 is divisible by 72, then what is the value of (7x + 2y), for the maximum value of y?
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The number 3798125P369 is divisible by 7. What is the value of the digit P?
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Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.
select the correct answer using the code given below:
