Question

How many numbers from $1$ to $10000$, inclusive, are multiple of exactly one of the numbers $3,5$ or $7$? For example,$28$ is such a number since it is a multiple of $7$ and not of $3$ and $7$ but $42$ is not such a number, since it is a multiple of both$3$ and $7$.

Answer 1

In order to determine the numbers of terms from $1$ to$10000$. The sequence of integers starting from $1$ to $10000$ is given by $1,2,3,4,.....,10000$. There are 1000 terms in the above sequence. The first and last terms are $1$ and $10000$ respectively and the common difference is $1$. Knowing the first and last term of an integer sequence, as well as the number of terms, we can calculate the sum of the first $n$ terms from this formula for the arithmetic sequence ${t_n} = a + (n - 1)d$.

Complete step by step solution:
We are given the numbers from $1$ to $10000$ that are multiple of exactly one of the numbers $3,5$ or $7$.
If all the terms of a progression except the first one exceeds the preceding term by a fixed number, then the progression is called arithmetic progression. If $a$ is the first term of a finite ${\rm A} \cdot {\rm P}$ and $d$ is a common difference, then ${\rm A} \cdot {\rm P}$ is written as $a,a + d,a + 2d \ldots \ldots \ldots ,a + \left( {n - 1} \right)d$
We take inclusive of the numbers from $1$ to $10000$ be multiplied by $3$ only. So we get
$3,6,9................9999$
We can calculate the total number of terms by the formula, ${t_n} = a + (n - 1)d$
Where, $a$ is the first term and ${t_n}$ be the last term and $d$is a common difference.
$9999 = 3 + (n - 1)3$ .since ${t_n} = 9999,a = 3,d = 3$
Expand the term into LHS, we can get

$$\dfrac{{9999 - 3}}{3} = n - 1 \\\ \dfrac{{9996}}{3} = n - 1 \\\$$

Simplify and finding the total number multiplied by $3$ only

$$3332 + 1 = n \\\ n = 3333 \to (1) \\\$$

We take inclusive of the numbers from $1$ to $10000$ be multiplied by $5$ only. So we get
$5,10,................10000$
We can calculate the arithmetic series of terms, ${t_n} = a + (n - 1)d$. Where, $a$ is the first term and ${t_n}$ be the last term and $d$ is a common difference
$10000 = 5 + (n - 1)5$.${t_n} = 10000,a = 5,d = 5$
Expand the term into LHS, we can get

$$\dfrac{{10000 - 5}}{5} = n - 1 \\\ \dfrac{{9995}}{5} = n - 1 \\\ 1999 + 1 = n \\\ n = 2000 \to (2) \\\$$

Simplify and finding the total number multiplied by $5$ only
We take inclusive of the numbers from $1$ to $10000$ be multiplied by $7$ only. So we get
$7,14,21................9996$
We can calculate the arithmetic series of terms, ${t_n} = a + (n - 1)d$ .Where, $a = 7$ is the first term and ${t_n} = 9996$ be the last term and $d = 7$is a common difference
$9996 = 7 + (n - 1)7$
Expand the term into LHS, we can get

$$\dfrac{{9996 - 7}}{7} = n - 1 \\\ \dfrac{{9989}}{7} = n - 1 \\\$$ $$1427 + 1 = n \\\ n = 1428 \to (3) \\\$$

We take inclusive of the numbers from $1$ to $10000$ be multiplied by $3\& 5$only.
Take LCM of $5\& 3$ is $15$
$15,30,................9990$
We can calculate the arithmetic series of terms, ${t_n} = a + (n - 1)d$. Where, $a = 15$ is the first term and ${t_n} = 9990$ be the last term and $d = 15$ is a common difference
$9990 = 15 + (n - 1)15$
Expand the term into LHS, we can get

$$\dfrac{{9990 - 15}}{{15}} = (n - 1) \\\ \dfrac{{9975}}{{15}} = (n - 1) \\\$$ $$665 + 1 = n \\\ n = 666 \to (5) \\\$$

$\Rightarrow$ We take inclusive of the numbers from $1$to$10000$ be multiplied by $5\& 7$only. We can get,
Take LCM of $5\& 7$ is $35$
$35,70,................9975$
We can calculate the arithmetic series of terms, ${t_n} = a + (n - 1)d$ .Where, $a = 35$ is the first term and ${t_n} = 9975$ be the last term and $d = 35$ is a common difference
$9975 = 35 + (n - 1)35$
Expand the term into LHS, we can get

$$\dfrac{{9975 - 35}}{{35}} = (n - 1) \\\ \dfrac{{9940}}{{35}} = (n - 1) \\\$$ $$284 + 1 = n \\\ n = 285 \to (6) \\\$$

We take inclusive of the numbers from $1$ to $10000$ be multiplied by $7\& 3$ only. We can get
Take LCM of $3\& 7$ is $21$
$21,42,................9996$
We can calculate the arithmetic series of terms, ${t_n} = a + (n - 1)d$. Where, $a = 21$ is the first term and ${t_n} = 9996$ be the last term and $d = 21$is a common difference
$9996 = 21 + (n - 1)21$
Expand the term into LHS, we can get

$$\dfrac{{9996 - 21}}{{21}} = (n - 1) \\\ \dfrac{{9975}}{{21}} = (n - 1) \\\$$ $$475 + 1 = n \\\ n = 476 \to (6) \\\$$

$\Rightarrow$ We take inclusive of the numbers from $1$ to $10000$ be multiplied by $3,7,5$ only. We can get
Take LCM of $3,5,7$ is $105$
$105,210,................9975$
We can calculate the arithmetic series of terms, ${t_n} = a + (n - 1)d$ .Where, $a = 105$ is the first term and ${t_n} = 9975$ be the last term and $d = 105$ is a common difference
$9975 = 105 + (n - 1)105$
Expand the term into LHS, we can get

$$\dfrac{{9996 - 105}}{{105}} = (n - 1) \\\ \dfrac{{9870}}{{105}} = (n - 1) \\\$$ $$94 + 1 = n \\\ n = 95 \to (7) \\\$$

Now, we have to calculate the total number that is only multiplied by $3,5,7$ . The remaining terms will be subtracted from this term as follows.
We can calculate the sum of the terms equation $(1),(2),(3)$ but subtract $(4),(5),(6)$ and $(7)$, we get

$$\Rightarrow (1) + (2) + (3) - (4) - (5) - (6) - (7) \\\ \Rightarrow 3333 + 1428 + 2000 - 666 - 285 - 476 - 95 = 5239 \\\$$

Therefore, $5239$ is the numbers from $1$ to $10000$, inclusive are multiples of exactly one of the numbers $3,5$ or $7$.

Note:
We note that the first and last term of the arithmetic sequence can be calculated from the number $1$ to $10000$. The arithmetic sequence formula is ${t_n} = a + (n - 1)d$. Finally we get the $n$ terms of the given problem. Then we perform the summation of all values then find the appropriate number.