1. ## question4

x and y are positive integers, There isn't 0 digit in both of them(x and y).

if $x.y=10.000$

What is different of $x-y=?$

a)609 b)631 c)643 d)789 e)919

2. Originally Posted by OPETH
x and y are positive integers, There isn't 0 digit in both of them(x and y).

if $x{\color{red}.} y=10{\color{red}.} 000 x-y=?$

a)609 b)631 c)643 d)789 e)919
I . meant to be a decimal point or is it being used as multiplication on the left side. Your notation has caused confusion on at least one other occassion.

3. Originally Posted by mr fantastic
I . meant to be a decimal point or is it being used as multiplication on the left side. Your notation has caused confusion on at least one other occassion.

Can you look at the question4 again,please?

4. Originally Posted by OPETH
Can you look at the question4 again,please?
Will you please answer the question about what you mean by $x.y$

5. ${5}^4= 625$

${2}^4= 16$

$625-16= 609$

$625.16=10 000$ or $5^4.2^4=10 000$

6. Originally Posted by OPETH
$x.y=10.000$
They want to know what you meant by $10.000$ but I'll just assume you meant plain old $10000$

On a side note, just use $xy$ instead of $x.y$ or at the very least try $x\cdot y$

There is no shortcut in this problem, you're just going to have to break up 10000 into it's factors, find the pair that doesn't have any zeros and then find the difference.

EDIT: Now I've noticed you've finished the problem already

7. Hello, OPETH!

$x$ and $y$ are positive integers; neither contains a zero.

If $x\cdot y\:=\:10,000$, what is: $|x-y|$ ?

. . $a)\;609\qquad b)\;631\qquad c)\;643\qquad d)\;789\qquad e)\;919$

The factors of 10,000 are: . $\begin{array}{ccc}1&\times&10,000 \\ 2&\times&5,000 \\ 4&\times&2,500 \\ 5&\times&200 \\ 8&\times&1250 \\ 10&\times&1,000 \\ {\color{blue}16}&\times&{\color{blue}625} \\ 25&\times&400 \\ 40&\times&250 \\ 50&\times&200 \\ 80&\times&125 \\ [-3mm]\end{array}$
. . . . . . . . . . . . . . . . . . . $\begin{array}{ccc}100& \times & \;\;100 \end{array}$

The only pair of factors without a zero is: . $16\cdot625$

Therefore: . $625 - 16 \:=\:\boxed{609}\quad\hdots \text{ answer (a)}$

8. nice one, but how did you get all the factors of 10,000

Did you have to manually go through and check, or use a table? In an exam situation this would be tricky without a table, there’s no way you would have time to work out the factors, how did you do this?

9. Sorry dear If you didnt have the possible answers ,this problem was not solvable.

10. Originally Posted by Maths_Tutoring
nice one, but how did you get all the factors of 10,000

Did you have to manually go through and check, or use a table? In an exam situation this would be tricky without a table, there’s no way you would have time to work out the factors, how did you do this?
You can get "all" the factors of any number by looking at its "prime factorization":
10000 is even so 2 is a factor and dividing by 2 leaves 5000. That is also even so 2 is still a factor and dividing by 2 again leaves 2500. That is also even so 2 is still a factor and dividing by 2 again leaves 1250. That is also even so 2 is still a factor and dividing by 2 again leaves 625. That is NOT even so 2 there is no factor of 2. Dividing by 3 does not give an whole result so 3 is not a factor. Dividing by 5 gives 125 so 5 is a factor. Dividing that by 5 again gives 25 so 5 is again a factor. Dividing by 5 gives 5:

There are 4 factors of 2 and 4 factors of 5: (2^4)(2^4).

That will work for any number. 10000 happens to be very simple and I suspect that Soroban just saw immediately that 10000= (10)^4 (because there are 4 0's) =(2*5)^4= (2^4)(5^4).

Once we have that, we can construct all ways of multiplying TWO numbers to get 10000 by combing those factors in different ways.

However, in this problem we are told that "There isn't 0 digit in both of them(x and y)". Since 2*5= 10 and a factor of 10 will always give a 0 digit, we cannot multiply and 2's or 5's together. The only possible way to avoid having 2's and 5's in the same factor is to have 2^4= 16 and 5^4= 25 as the two factors.

OPeth, if you are still with us, when someone asks a question about your post, answer the question; don't just tell them to look at the post again!