the last digit

[Calculator]

What is the last digit of the product of numbers: 1x3x5x7...x99 ?
A friend told me to use the factorials, but we still haven't covered that section, so is there an easier way to calculate this?

 

[Quaid]

Here's one approach.

First, multiplication by 1 does not change anything, so we can drop that factor.

Second, consider reversing the order of multiplications.

99 * 97 * 95 * ... 7 * 5 * 3

Now, consider just the part in red. You multiply a bunch of odd numbers, the last of which is 5.

If you multiply any non-zero number by 5, what is the last digit of the product? In other words, what does the last digit need to be, in any multiple of 5 (ignoring zero)?

Finally, consider the factor of 3. If you start with a multiple of 5, and you multiply that number by 3, does the last digit change?

If you're not sure, pick some multiples of 5 and multiply them by 3 longhand. See what happens, to generate the last digit of the product.



PS: Beginning with algebra, we stop using the letter x as a multiplication symbol. Use an asterisk or grouping symbols, instead.

 

[Deleted member 4993]

To emphasize Quaids point:

1 * 3 * 5 = 15

1 * 3 * 5 * 7 = 105

1 * 3 * 5 * 7 * 9 = 945

 

[lookagain]

.
An odd number can be represented by 10b+1, where b is any integer.

I wanted to clarify this. An infinite number of odd numbers can be represented by 10b + 1,
where b is any integer, but not all of the odd numbers.

 

[Calculator]

Thank you all

 

[HallsofIvy]

A number ending with 5 can be represented by 10a+5, where a is any integer.
An odd number can be represented by 10b+1, where b is any integer.

(10a+5)(10b+1) = 100ab + 50b + 10a + 5 : get my drift?

I suspect that Denis meant to say "An odd number can be represented by 2b+ 1 where b is any integer"

Then (10a+ 5)(2b+ 1)= 20ab+ 10b+ 10a+ 5= 10(2ab+ b+ a)+ 5 making the same point.

 

[HallsofIvy]

Very intelligent of you.
(If it were me, I would have forgotten what I meant by this time.)