7.1. On the counter lie 10 weights with masses $n, n+1, \ldots, n+9$. The seller took one of them, after which the total weight of all the remaining ones was 1457. Which weight did the seller take?
# Answer: 158

Solution. The total weight of all 10 weights is $10 n+45$. Suppose the weight taken has a weight of $n+x$, where $x<10$. Then

$10 n+45-n-x=1457$.

$9 n+45-x=1457$.

We can see that 1457, when divided by 9, gives a remainder of 8, so $x=1$. Then

$9 n+45-1=1457$

$9 n=1413$.

$n=157$.

$n+x=158$.