Stupefy rennervate

We can simplify the fraction $\frac{12}{21}$ as $\frac{4}{7}$ . Therefore, $a$ and $b$ must be $4x$ and $7x$ respectively, assuming that $x$ is a positive integer. We can solve this problem by finding the number of possible values of $x$ there is.

Since $a=4x$ , $4x$ must be at least 100 and at most 1000. The same applies for $b$ , which is equivalent to $7x$ . Let us think about the lower boundary first (being at least 100). We can say that $4x \leq 100$ , meaning that $x \leq 25$ . Therefore, the lowest possible value of $x$ is $25$ . (Note that we did not need to use $7x \leq 100$ , because then, some of the values of $x$ would not show $4x \leq 100$ to be true.) For the upper boundary, $7x \leq 1000$ , which makes $x \leq 142$ (remember $x$ must be a positive integer). (Note again that we do not need to use $4x \leq 1000$ , because for some values of $x$ , the inequality $7x \leq 1000$ will not be true.) So, the range of $x$ is from $25$ to $142$ . Simple arithmetic leads us to our final answer: $142-25+1=\boxed{118}$

We're looking for pairs of integers a and b such that : $\frac{a}{b} = \frac{12}{21} = \frac{4}{7}$

We know that every fraction equal to $\frac{4}{7}$ can be written : $\frac{4\times k}{7\times k} , k \in Z$

So, if $\frac{a}{b} = \frac{4}{7}$ , then : $a = 4k$ $b = 7k$ $k \in Z$

Since : $100 \leq a \leq 1000 \\ 100 \leq b \leq 1000$ $\Leftrightarrow$ $100 \leq 4k \leq 1000 \\ 100 \leq 7k \leq 1000$ $\Leftrightarrow$ $25 \leq k \leq 250 \\ 15 \leq k \leq 142$ $\Leftrightarrow$ $25 \leq k \leq 142$

There are exactly $(142 - 25) + 1 = 118$ possible values for $k$ . So there are 118 pairs of integers a and b .

We see that $\dfrac{a}{b}=\dfrac{4}{7}$ Any $a$ and $b$ satisfying this is a solution. The lowest $a$ is 100, when $b$ is 175. The highest $b$ is 994. This is $\dfrac{994-175}{7}+1=\boxed{118}$ ordered pairs.

$\frac{a}{b}=\frac{12}{21}=\frac{4}{7}$ . From this, we can say that the smallest possible value of $a$ is 100, which is the 25th positive multiple of 4. Now, the highest possible value of $b$ under the given conditions is 994, and this number is the 142nd positive multiple of 7 ( $\frac{994}{7}=142$ ). Therefore, the number of possible pairs of integers $(a,b)$ is 142 - 25 + 1 = 118.

$\frac {12}{21} = \frac {4}{7}$ . We construct the equations: $4m \geq 100$ and $7n \leq 1000$ with the purpose of looking for the minimum and maximum value that can multiply the $4$ and $7$ (respectively), satisfying the condition of being between $100$ and $1000$ . So, we get $m \geq 25$ and $n \leq 142.86$ (rounded). As $a$ and $b$ are integers, and saying that $4$ and $7$ will get multiplied by some $k$ , then $25 \leq k \leq 142$ . Finally, from $25$ to $142$ there are $142 - 25 + 1 = 118$ integers. Thus, the answer is $\boxed{118}$ .

We see that a/b=4/7 Any a and b satisfying this is a solution. The lowest a is 100, when b is 175. The highest b is 994. This is [(994−175)/7]+1=118 ordered pairs.

$a:b=12:21=4:7$

$a|4$ and $b|7$ .

Since $100 \leq a \leq 1000$ ,the minimum value of $a=25\times4=100$ .Then $b=100 \times \frac{7}{4}=175$ .

$100 \leq b \leq 1000$ ,so the maximum value of $b=142\times7=994$ .Then $a=994 \times \frac{4}{7}=568$ .

$100 \leq a \leq 568$

$175 \leq b \leq 994$

To find the number of pairs $(a,b)$ ,we just need to count the number of possibilities of $a$ .

$a$ can be any multiple of 4 from $100$ to $568$ ,so the number of possibilities

= $\frac{568-100}{4}+1=\frac{468}{4}+1=118$

Hence,the number of pairs $(a,b)$ = $\boxed{118}$

a/b=12/21=4/7 implies a=4b/7 and b=7a/4. Since a and b both must be integers, b must be a multiple of 7 falling between 100 and 1000 which would also give a value of a that lies in 100 to 1000. Similarly, a must be a multiple of 4. As a is always smaller than b, we will check the lower limit for a and upper limit for b. The smallest multiple of 4 in given range is 100, which corresponds to (100,175). The largest ,multiple of 7 in given range is 994. Thus, now we only have to count multiples of 7 from 175 to 994, which can be done easily. Answer comes out to be 118.

$\frac{a}{b} = \frac{4}{7}$ let $a:b:k=4:7:1$ minimum value of k is $ceil(max(\frac{a_{min}}{4},\frac{b_{min}}{7}))=ceil(max(\frac{100}{4},\frac{100}{7}))=ceil(\frac{100}{4})=25$ maximum value of k is $floor(min(\frac{a_{max}}{4},\frac{b_{max}}{7}))=floor(min(\frac{1000}{4},\frac{1000}{7}))=floor(\frac{1000}{7})=142$ k has possible values between 25 and 142, inclusive. There are (142-25+1)=118 possible values

since a/b=12/21=4/7 so a values are divisible by 4, and b values are divisible by 7. since a<b, so the highest maximum common divisor would calculated from b, and the lowest maximum common divisor would calculated from a the highest maximum common divisor = 1000/7=142.8=142 the lowest maximum common divisor= 100/4=25 so the total numbers of pairs = the numbers from 25 to 142, each are included=142-24=118

a/b = 12/21 So, a/b=4/7 S, b=7a/4....... if a=100 then b=175 (100,175) To find the number of ordered pairs we need to know how many multiples of 7 are there between 175 and 1000. (1000-175)/7=117.8 So, there are 117 + 1 = 18 multiples of 7 between 175 and 1000 including the 175 (that's why we must add 1). Therefore, there are 118 pairs of (a,b).