What is the remainder when
\(70 \times 71 \times 72 \times 73 \times 74 \times 75 \times 76 \times 77 \times 78 \times 79\) is divided by 1000 ?
Remainder Problem Analysis
The question asks us to find the remainder when the product of ten consecutive integers, starting from 70 up to 79, is divided by 1000. The product is \(P = 70 \times 71 \times 72 \times 73 \times 74 \times 75 \times 76 \times 77 \times 78 \times 79\). We need to calculate \(P \pmod{1000}\).
Divisibility Rules Strategy
To find the remainder when dividing by 1000, we can analyze the divisibility of the product \(P\) by the prime factors of 1000. First, let's find the prime factorization of 1000.
\(1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3\)
So, \(1000 = 8 \times 125\).
If the product \(P\) is divisible by both 8 and 125, it will also be divisible by their product, 1000, because 8 and 125 share no common factors other than 1 (they are coprime).
125 Divisibility Check
We need to determine if the product \(P\) has enough factors of 5 to be divisible by \(5^3 = 125\). Let's examine the numbers in the sequence: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79.
We look for multiples of 5 within this range:
- 70: \(70 = 14 \times 5 = (2 \times 7) \times 5\). This number contributes one factor of 5.
- 75: \(75 = 15 \times 5 = (3 \times 5) \times 5 = 3 \times 5^2\). This number contributes two factors of 5.
The total number of factors of 5 in the product \(P\) is the sum of the factors from 70 and 75, which is \(1 + 2 = 3\).
This means the product \(P\) contains \(5^3\) as a factor. Therefore, \(P\) is divisible by 125.
8 Divisibility Check
Next, we check if the product \(P\) has enough factors of 2 to be divisible by \(2^3 = 8\). We examine the even numbers in the sequence: 70, 72, 74, 76, 78.
- 70: \(70 = 2 \times 35\). This contributes one factor of 2 (\(2^1\)).
- 72: \(72 = 8 \times 9 = 2^3 \times 9\). This contributes three factors of 2 (\(2^3\)).
- 74: \(74 = 2 \times 37\). This contributes one factor of 2 (\(2^1\)).
- 76: \(76 = 4 \times 19 = 2^2 \times 19\). This contributes two factors of 2 (\(2^2\)).
- 78: \(78 = 2 \times 39\). This contributes one factor of 2 (\(2^1\)).
The total number of factors of 2 in the product \(P\) is at least \(1 + 3 + 1 + 2 + 1 = 8\).
Since the product \(P\) contains \(2^8\) as a factor, and \(8 \ge 3\), \(P\) is definitely divisible by \(2^3 = 8\).
Conclusion: Remainder Calculation
We have found that the product \(P\) is divisible by 125 (since it contains \(5^3\)) and also divisible by 8 (since it contains \(2^8\), which is more than \(2^3\)).
Because \(P\) is divisible by both 8 and 125, and these two numbers are coprime, \(P\) must be divisible by their product, \(8 \times 125 = 1000\).
Therefore, the remainder when \(70 \times 71 \times 72 \times 73 \times 74 \times 75 \times 76 \times 77 \times 78 \times 79\) is divided by 1000 is 0.
Similar Questions
- If \(n\) is a natural number, then what is the sum of all distinct remainders of \(4^n + 6^n + 9^n + 11^n\) when divided by 10 for various values of \(n\)?
- If \(a, b, c, d\) are natural numbers, then how many possible remainders are there when \(1^a + 2^b + 3^c + 4^d\) is divided by 10?
- The expression \(555^{777} + 777^{555}\) is divisible by which of the following?
1. 2
2. 3
3. 37
Select the correct answer using the code given below: - When the number (12345678910111213 ... 99100) is divided by 16, what will be the remainder?
- What is the smallest natural number \(n\) such that \((n + 1) \times n \times (n - 1) \times (n - 2) \times ... 3 \times 2 \times 1\) is divisible by 910?
- Consider the following statements :
1. If \((3m^3 + 2m^2 + 5m + n)/m\) is not an integer, where \(m\) and \(n\) are integers, then \(n\) is not divisible by \(m\).
2. \(5(8^m) + 2^{3m}\) is divisible by 48 for all whole numbers \(m\).
Which of the statements given above is/are correct? What is the remainder when \(111^{222} + 222^{333} + 333^{444}\) is divided by 5?
What is the remainder when \(3^{255}\) is divided by 28?
What is the value of \(x (0 \le x \le 8)\) if \((100^{97} + 100^{54} + x + 1)\) leaves a remainder 0 when divided by 9?
- What is the remainder if we divide \(3^{10}\) by 7?
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?
If all positive divisors of 132 are arranged in descending order, then what digit will be at unit place of first divisor ?
If 3 2019 is divided by 10, then what is the remainder?
The number 3798125P369 is divisible by 7. What is the value of the digit P?
Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
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:
