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.
Important Questions from Divisibility and Remainder
What is the sum of the digits of the least number which when divided by 12, 16 and 20 leaves the same remainder 6 in each case and it is divisible by 9?
As nine-digit number 89563x87y is divisible by 72. What is the value of \(\sqrt{7x-3y}\) ?
The greatest number that on dividing 2675 and 2320 leaves the reminder 5 and 6 ,respectively is :
Find the greatest number that exactly divides 2880, 6525 and 8307.
If a 10 - digit number 643x1145y2 is divisible by 88, then the value of (2x - 3y) for the largest value of y is :
