Cool Problem

Number Theory Level 3

Let $f(n)$ denote the sum $1+2+3+...+(n-1)+n$ . Let $g(n)$ denote the sum $1+3+...+k$ , where $k$ is the $n$ th positive odd number. There are $x$ integers n such that $1≤n≤100$ and $2f(n)-g(n)$ is divisible by both $3$ and $2$ . What is the sum of the digits of $x$ ?

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3 5 10 7

1 solution

Ivan dennys Opit
Jan 2, 2015

2f(n)=n\times{n+1) and g(n)=n^2 then 2f(n)-g(n)=n^2+n-n^2=n and n is divisible by 3 and 2 n is 6x, because n<100 and 6\times 16=96 so there are 16 integers x satisfy n value finally the sum digits is 1+6=7

Nice problem with nice solution :)

Mahtab Hossain - 6 years ago

Cool Problem

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