Question

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.
(b) are divisible by 7.
(c) are even.
(d) are divisible by 5 but not by 7.

Answer 1

Answer:

(a) There are 257 numbers which are divisible by 5 and by 7

(b) There are 1286 numbers which are divisible by 7

(c) There are 4500 even numbers

(d) There are 514 numbers which are divisible by 5 but not by 7

Step-by-step explanation:

(a) Since, LCM(5, 7) = 35,

Thus, the numbers divisible by 5 and by 7 between 1000 and 9999,

1015, 1050, 1085............, 9975

Which is an AP,

Having first term, a = 1015,

Common difference, d = 35,

Last term, l = 9975

If n be the number of terms,

$\because l = a + (n-1)d$

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

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

$8960 = (n-1)35$

$256 = n-1$

$\implies n = 256 + 1 = 257$

(b) Similarly,  number divisible by 7,

1001, 1008, 1015............, 9996

Here, a = 1001, d = 7, l = 9996

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

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

$8995 = (n-1)7$

$1285 = n-1$

$\implies n = 1285 + 1 = 1286$

(c) Even numbers between 1000 and 9999 inclusive

1000, 1002, 1004, 1006,........9998

Here, a = 1000, d = 2, l = 9998

$9998 = 1000 + (n-1)2$

$9998 - 1000 = (n-1)2$

$8998 = (n-1)2$

$4499 = n-1$

$\implies n = 4499 + 1 = 4500$

(d) Number divisible by 5,

1000, 1005, 1010............, 9995

Here, a = 1000, d = 5, l = 9995

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

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

$8995 = (n-1)5$

$1799 = n-1$

$\implies n = 1799 + 1 = 1800$

Hence, the number divisible by 5 but not by 7 = numbers divisible by 5 - number divisible by 7

= 1800 - 1286

= 514