Question

(1 point) Find how many positive integers with exactly four decimal digits, that is, positive integers between 1000 and 9999 inclusive, have the following properties: (a) are divisible by 5 and by 7. 257 (b) are divisible by 5 but not by 7. (c) are divisible by 5. (d) are divisible by 7.

Answer 1

Using arithmetic sequences, it is found that there are:

a) 257 numbers that are are divisible by 5 and by 7.

b) 1543 numbers that are divisible by 5 but not by 7.

c) 1800 numbers that are divisible by 5.

d) 1286 numbers that are divisible by 7.

In an arithmetic sequence, the difference between consecutive terms is always the same, called common difference d.

The nth term of an arithmetic sequence is given by:

$a_n = a_1 + (n-1)d$

In which $a_1$ is the first term.

Item a:

• Numbers that are divisible by 5 and 7 are multiples of 35, as the least common multiple of 5 and 7 are 35, thus we have an arithmetic sequence with $d = 35$.
• The first multiple of 35 in this interval is 1015, so $a_1 = 1015$.
• The last multiple of 35 in this interval is 9975, so $a_n = 9975$.

We apply the general formula and solve for n, thus:

$a_n = a_1 + (n-1)d$

$9975 = 1015 + 35(n-1)$

$35n - 35 = 8960$

$35n = 8995$

$n = \frac{8995}{35}$

$n = 257$

There are 257 numbers that are are divisible by 5 and by 7.

Item b:

First, we find the number of multiples of 5, with $a_1 = 1000, a_n = 9995, d = 5$. Thus:

$a_n = a_1 + (n-1)d$

$9995 = 1000 + 5(n-1)$

$5n - 5 = 8995$

$5n = 9000$

$n = \frac{9000}{5}$

$n = 1800$

Of those numbers, 257 are multiples of 35, which means that they also are multiplies of 7, thus they are removed.

1800 - 257 = 1543.

There are 1543 numbers that are divisible by 5 but not by 7.

Item c:

From the previous item, there are 1800 numbers that are divisible by 5.

Item d:

We find the number of multiples of 7, with $a_1 = 1001, a_n = 9996, d = 7$. Thus:

$a_n = a_1 + (n-1)d$

$9996 = 1001 + 7(n-1)$

$7n - 7 = 8995$

$7n = 9002$

$n = \frac{9002}{7}$

$n = 1286$

There are 1286 numbers that are divisible by 7.

A similar problem is given at https://brainly.com/question/23901992