All the numbers from 1 to 99 are multiplied together. Use a pattern to determine the last digit of the product. Justify your answer.
1 = 1 = 1!
1*2 = 2 = 2!
1*2*3 = 6 = 3!
1*2*3*4 = 24 = 4!
1*2*3*4*5 = 120 = 5!
1*2*3*4*5*6 = 720 = 6!
...
I assume that 99! will be even and will also end in a 0
since 99! = 1*2*3*4*5*...
2*5 is a factor, so 99! is a multiple of 10, ending in zero.
To determine the last digit of the product of all the numbers from 1 to 99, we need to look for patterns in the multiplication of the last digits.
Here's a step-by-step explanation:
1. Write down the last digit of each number from 1 to 99.
1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, ...
2. Look for patterns in the last digits as we multiply them together.
1 * 2 = 2
2 * 3 = 6
6 * 4 = 24
4 * 5 = 20
0 * 6 = 0
As we continue this pattern, we observe that when we multiply any number ending in 4 or 6 with an even number, the result will end in 0. Similarly, whenever we multiply any number ending in 2 or 8 with an odd number, the result will also end in 6.
This pattern will continue throughout the multiplication.
3. Based on the above observation, we can conclude that the product of all the numbers from 1 to 99 will end with the digit 0.
Therefore, the last digit of the product of all the numbers from 1 to 99 is 0.