how many integer between 1 and 566 are divisible by either 3 and5
Answers
Answer: Number of integer between 1 and 566 are divisible by either 3 and 5 is 264
Step-by-step explanation:
nth term of an AP with first term 'a' and common difference 'd' is given by
nth term = a + (n-1)d
To find the number of integers divisible by 3
Let first term 'a' of AP is 3 and last term (nth) is 564
Using nth term equation
nth term = a + (n-1)d
3 + (n- 1)3 = 564
n = 188
There are 188 terms which are divisible by 3
To find the number of integers divisible by 5
Let first term 'a' of AP is 5 and last term (nth) is 565
Using nth term equation
nth term = a + (n-1)d
5 + (n- 1)5 = 565
n = 113
There are 188 terms which are divisible by 3
To find the number of integers divisible by 3 and 5
For this we find number of integers divisible by 15
Let first term 'a' of AP is 15 and last term (nth) is 555
nth term = a + (n-1)d
15 + (n- 1)15 = 555
n = 37
There are 188 terms which are divisible by 3
Therefore number of integer between 1 and 566 are divisible by either 3 and5 = 188 + 113 - 37 = 264 numbers
Answer:
Number of integers are 264
Step-by-step explanation:
Given, integer b/w 1 and 566
Let consider a Arithmetic Progression (AP) series. Now, nth term of an AP with first term 'a' and common difference 'd' is given by
nth term = a + (n-1)d
Condition 1: To find the number of integers divisible by 3
Let a=3 and last term is 564 and d=3
Using nth term equation
nth term = a + (n-1)d
3 + (n- 1)3 = 564
3+ 3n-3=564
3n=564
n= 188
Condition 2: To find the number of integers divisible by 5
Let a= 5 and last term is 565
Using nth term equation
nth term = a + (n-1)d
5 + (n- 1)5 = 565
n = 113
Condition 3: As some of integers are common therefore it is necessary to find number of integers divisible by 3 and 5 i.e. 15
Let a= 15 and last term is 555
nth term = a + (n-1)d
15 + (n- 1)15 = 555
n = 37
Therefore, number of integer between 1 and 566 divisible by either 3 and 5 = 188 + 113 - 37 = 264