when a number is divided by 15 the remainder is 7 and when it is devied by 21 it gives reminder 10.Find how many such number exist in between 80 to 7000.
Answers
Given: a number is divided by 15 the remainder is 7 and when it is devied by 21 it gives reminder 10.
To find : how many such number exist in between 80 to 7000.
Solution:
when a number is divided by 15 the remainder is 7
=> N = 15a + 7
when it is devied by 21 it gives reminder 10
=> N = 21b + 10
15a + 7 = 21b + 10
=> 15a = 21b + 3
=> 5a = 7b + 1
b = 2 , a = 3
b = 7 , a = 10
b = 12 , a = 17
15 & 21 have LCM as 105
=> N = 52 , 157 , 262 .........
difference of 105 in consecutive number
52 < 80
157 is 1st
let say n such numbers
157 + (n - 1) 105 < 7000
=> (n - 1) < 65.17
=> n < 66.17
=> n = 66
66 such numbers between 80 to 7000
157 , 262 ................................................ 6982
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