Question

alec and beau race to compute $1 2 3 \cdots n$. alec skips two numbers and gets a sum of $92$. beau double-counts two numbers and gets a sum of $145$. what is the value of $n$?

Answer 1

Hi,

n=15

Explanation:

Imagine that Alec have not skiped 2 numbers.
`1+2+...+n=92 =\frac{n*(n+1)} {2}\\n^2+n=2*92\\n^2+n-184=0\\\\n=(-1+√(729) )/(2) =13\\or\\n=(-1-√(729) )/(2) =-14\ (hence\ n > 0)\\\\n=13\\`

In reality, n=13+2=15 an Alec has skiped 13 and 15

1+2+3+...+15=15*16/2= 120

120-92=28

a+b=28 and a,b ≤15 ==> 28=13+15

Beau:

145=120+25

25=15+10=14+11=13+12

Beau has double- counts (15,10) or (14,11) or (13,12)