Question

How many positive integers are these 0 to 1000 that leave a remainder of 3 on division by 7and a remainder of 2 on division by 4?

Answer 1

Step-by-step explanation:

Series of no. which leave remainder 3 on division by 7 is :

10 , 17 , 24 , 31 , 38 , 45, 52 , 59 , 66 ,71 , …… so on …………(let it be series “A”)

-series of no. which leaves remainder 2 on division by 4 is:

6 , 10 , 14 , 18 , 22 , 26 , 30 , 34 , 38 , 42 , 46 ,50 , 54 , 58 , 62 , 66……so on ………(let it be series “B”).

Now the common terms between the series “A” and “B” is the required series which leaves remainder 3 when divided by 7 and remainder 2 on division by 4 .

so the required series is

10 , 38 , 66…. so on

Now by logic we can imagine, the above series will be in AP with first term as 10 and common difference as “28”

No we just have to find the total no. of term in the above series with last term less then or equal to 1000.

so using formula , last term = 10 + (n-1)28

let last term be 1000

so , 1000 = 10 + (n-1)28

solving ,

n = (1000 - 10)/28 + 1 ,

n= 36.35..

since “n” is integer and last term should be less then “1000”

hence n= 36 is the only possible no. of terms

check :

last term = 10+ (36–1) 28 …………(using formula of AP ‘s last term)

last term = 990

now u can easily see that 990 is the last no. in the series which leave remainder 3 when divided by 7 and remainder 2 when divided by 4.

hence the complete series is 10 , 38, 66 , …., 990 .