Lazy Liz divide and conquer

$\frac{a}{b} \div \frac{c}{d}= \frac{a}{b} \times \frac{d}{c} =\frac{a \times d}{b \times c}$ by normal division. We can take this further, though. We first rearrange the terms to get $\frac{a \times d}{c \times b}$ Next, we split the multiplications to get $\frac{a}{c} \times \frac{d}{b}$ Rewriting $\dfrac{a}{c}$ as $a \div c$ , we get $(a \div c) \times \frac{d}{b}$ rewriting $\dfrac{d}{b}$ as $\dfrac{1}{\frac{b}{d}}$ gives $(a \div c) \times \frac{1}{ \frac{b}{d}}$

Finally, rewriting $\frac{b}{d}$ as $\dfrac{1}{b \div d}$ This gives $\frac{ a \div c}{ b \div d} .$ Thus, Lazy Liz's division is correct, so all that remains is counting the number of ways we can find the ordered quadruples without restrictions. This can be counted as $5 \times 5 \times 5 \times 5= \boxed{625}$

Lazy Liz is correct, so $a, b, c, d$ can all take on any of the values, giving $5^4=\boxed{625}$ possibilities.

Everything about this question is designed to mislead. I spent a fair bit of time puzzling over many counterexamples before I realised.

"Standard" division is taught in primary schools as "turn the second fraction over and multiply". We are really multiplying both fractions by the multiplicative inverse of the second fraction but the primary school definition will be fine for us.

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$

Now Liz wants to divide the two fractions like this instead:

$\frac{a}{b} \div \frac{c}{d} = \frac{a \div c}{b \div d} = \frac{a}{c} \div \frac{b}{d}$

Inverting the second fraction and multiplying gives:

$\frac{a}{c} \div \frac{b}{d} = \frac{a}{c} \times \frac{d}{b} = \frac{ad}{cb}$

Now multiplication is commutative so:

$\frac{ad}{bc} = \frac{ad}{cb}$

Although it takes more effort to compute, Liz' method always works. The number of correct values is just $5^{4}$ or $625$ using the rule of product.

The most valuable lesson from this question is that Brilliant question setters are horrible people, because divide and conquer is never used, because they marginalize Liz' views and because they create untrue stereotypes. For shame!

let's do some algebra $\frac{a}{b} \div \frac{c}{d} =\frac{\frac{a}{b}}{\frac{c}{d} }=\frac{ad}{bd},$ on the other hand
$\frac{a\div c}{b\div d}=\frac{\frac{a}{c}}{\frac{b}{d}} =\frac{ad}{bd},$ so $\frac{a}{b} \div \frac{c}{d} =\frac{a\div c}{b\div d}.$ Hence each integer has the posibility of been 1,2,3,4,5 so each one has 5 posibilities $5\times 5 \times 5 \times 5 \times 5 =\boxed{625}$

This one looks tricky, but it's still right! If you don't trust me, I'll prove it for you.

$\frac{a}{b} \div \frac{c}{d}$

$= \frac{a}{b} \times \frac{d}{c}$

$= \frac{ad}{bc}$

Multiply by $\frac{1}{cd}$ on num. and denom. we get...

$= \frac{a\frac{1}{c}}{b\frac{1}{d}}$

$= \frac{a \div c}{b \div d}$ . ~~~

If all numbers satisfy this equation (not really an equation.), then the number of quadruples $(a,b,c,d) = 5\times 5\times 5\times 5 = 5^{4} = \boxed{625}$ ~~~

By simplifying both sides into one fraction, we get $\frac{ad}{bc}$ on both sides. This means that no matter what numbers there are, the stated equation works. Because a, b, c, and d must all be an integer 1 thorough 5, there are 5 choices for a, 5 choices for b, etc. We multiply all of the choices to get: $5\cdot5\cdot5\cdot5=\boxed{625}$ different ordered quadruples of integers between 1 and 5 inclusive that satisfy this condition.

Actually, Lazy Liz's thinking is correct! :D

$\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}$ $=\frac{a}{c}\times\frac{d}{b}$ $=\frac{a\div c}{b\div d}$

The integers $a,b,c,d$ are between $1$ and $5$ (inclusive), so each of the integers have 5 options.

There are $4$ integers.Therefore the answer is $5^4=\boxed{625}$

the given condition is true for all a,b,c,d not equals to zero. a can be chosen in 5 ways (1or 2 or 3 or 4 or 5) similarly b, c, d can be chosen in five ways each therefore, number of ordered pairs=5 5 5*5=625 Ans

The key to solving this is really just realizing... Lazy Liz isn't actually wrong.

Give it a try yourself: (a/b) / (c/d) and (a/c) / (b/d)

If you try simplifying both, you get an interesting result:

(a/b)/(c/d) is the same as a/(bc/d), which is the same as ad/bc. Or:

(a/b)/(c/d) = a/(bc/d) = (ad)/(bc)

So now try it with the other one:

(a/c) / (b/d) = a/(bc/d) = ad/bc

So they both come out to ad/bc. This means the statement at the bottom of our equation is actually:

ad/bc = ad/bc

Which is true for -all- quadruples of integers. That means, you simply need to multiply out the number of possible integers: there's five possible a values, five possible b values, five possible c values, and five possible d values. 5 * 5 * 5 * 5 = 625. Viola!

Lady Liz's idea is correct to any values of a,b,c,d between 1 and 5 inclusive. Proof:

$\frac{a}{b} / \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{a}{c} \times \frac{d}{b} = \frac{a}{c} / \frac{b}{d} = \frac{a / c}{b / d}$

Because this is generally true for every case of values of a,b,c,d in range 1 to 5, each term can be any of the integers from 1 to 5.

Now, for a chosen value of a, there are 5 possible values for b. Hence the total possible value combinations of a and b are: $5 \times 5 = 25$

Similarly, for a fixed value of a and b, there are 5 possible values for c, hence the total possible combinations of a, b and c are: $25 \times 5 = 125$

From there, we can work out the answer to the problem, which is the number of total possible combinations of all a,b,c and d: $125 \times 5 = 625$

The very first thing that we should do to solve this problem, is to simplify both sides of the equation so that it will be easier to understand.

$\frac{a}{b} \div \frac{c}{d} = \frac{a \div c}{b \div d}$

We will start with the LHS (left-hand side). Note that division by a fraction is the same as multiplication by the reciprocal of a fraction, so we can simplify as follows:

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$

The RHS (right-hand side) can be simplified as well, if we multiply the numerator and the denominator of the fraction by $cd$ to cancel out the division.

$\frac{a \div c}{b \div d} = \frac{cd(a \div c)}{cd(b \div d)} = \frac{ad}{bc}$

Notice that the left hand side and the right hand side are both $\frac{ad}{bc}$ ! Thus, all $(a, b, c, d)$ between 1 and 5 (inclusive) satisfy the given equation.

Now we must find the amount of distinct $(a, b, c, d)$ s. The first variable, $a$ , can take 5 different values (1 to 5), and so can $b, c,$ and $d$ . To find the total amount of possibilities, we can multiply the individual possibilities of each variable. That is $5^4 = \fbox{625}$ .

Here , both LHS and RHS are already equal so a,b,c and d can take all values that means 1,2,3,4 and 5. So, total no.s of quadruples are 5^{4}=625.

The whole problem is a trick! Lazy Liz isn't lazy as a matter of fact! She actually is just dividing in a little more of a complicated way! But dividing none the less! Therefore any quadruples will work! So the answer is 5 5 5*5, which is 625.

Lazy Liz has definitely done the right thing. Because the operation is true for all real numbers. And in this case, the number is limited for only integer from 1 to 5, hence there are $5 \times 5 \times 5 \times 5 =625$ integers.

We observe that a/b÷c/d is equal to ad/bc

And (a÷c)/(b÷d) is also equal to ad/bc

Hence the given condition holds for any value except b=c=d=0

Therefore total number of cases are 5 * 5 * 5 * 5 = 625 as interval is [1,5]