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

Hint: 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.