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How many natural numbers are there between 1to 200 which they are exatly divisible by 3, 9 and 27 |
| Answer» NUMBER divisible by 33,6,9……………198Using the FORMULA of finding nth term in an Arithmetic Progressionan= a+ (N-1)d ….. (1)where an: nth term, a:first term n:number of TERMS d:common difference198=3+(n-1)3=> n=66Number divisible by 77,14,………………………196196=7+(n-1)7 [from (1)]n=28Number divisible by 2121,42…………………….189189= 21+ (n-1)21 [from(1)]n=9Number divisible by 3 or 7 [not by 21] = 66+28–9=85… AnsHope it helps! | |