need help with 1-to-9 multiplacation table

[mathchic12]

If you think of a 1-to-9 multiplication table, are there more odd or more even products? How can you determine the answer without counting?
Is this different from a 1-to-9 addition facts table? Explain.

 

[wjm11]

If you think of a 1-to-9 multiplication table, are there more odd or more even products? How can you determine the answer without counting?
Is this different from a 1-to-9 addition facts table? Explain.

The product of an even number with either an even or odd number is always even.

The product of two odd numbers is odd.

Can you take it from there?

 

[Deleted member 4993]

Re: Confused

If you think of a 1-to-9 multiplication table, are there more odd or more even products? How can you determine the answer without counting?
Is this different from a 1-to-9 addition facts table? Explain.

If you multiply

odd * odd = odd

odd * even = even

even * odd = even

even * even = even

However, when you add

odd + odd = even

odd + even = odd

even + odd = odd

even + even = even

So what do you think?

 

[soroban]

Hello, mathichic12!

If you think of a 1-to-9 multiplication table, are there more odd or more even products?
How can you determine the answer without counting?

Is this different from a 1-to-9 addition facts table? Explain.

\(\displaystyle \text{The numbers are: }\;1,2,3,4,5,6,7,8,9\)
. . \(\displaystyle \text{There are: }\:9 \times 9 \:=\:81\text{ possible products.}\)


\(\displaystyle \text{There are }5\text{ odd numbers and }4\text{ even numbers.}\)

\(\displaystyle \text{The only way to get an odd product is: }\:\text{(odd)} \times \text{(odd)}\)
. . \(\displaystyle \text{Hence, there are: }\:5 \times 5 \:=\: 25\text{ odd products.}\)

\(\displaystyle \text{The other 56 products are even.}\)

\(\displaystyle \text{Therefore, there are more }even\text{ products.}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{On an addition table, there are: }\:9 \times 9 \:=\:81\text{ possible sums.}\)


\(\displaystyle \text{There are two ways to get an odd sum:}\)

. . \(\displaystyle \begin{array}{cccc} \text{(even) + (odd):}& 4 \times 5 &=& 20\text{ ways.} \\ \text{(odd) + (even):} & 5 \times 4 &=& 20\text{ ways.} \end{array}\quad\Rightarrow\quad\text{ 40 odd sums}\)

\(\displaystyle \text{And the other 41 are even sums.}\)