how many number like $119$ How many 3-digits number has this property like $119$:
$119$ divided by $2$ the remainder is $1$ 
119 divided by $3$ the remainderis $2 $
$119$ divided by $4$ the remainder is $3$
$119$ divided by $5$ the remainder is $4$
$119$ divided by $6$ the remainder is $5$
 A: Note that
$$x \equiv 1 \mod 2 \\ x \equiv 2 \mod 3 \\ x \equiv 3 \mod 4 \\ x \equiv 4 \mod 5 \\ x \equiv 5 \mod 6$$
is equivalent to
$$x \equiv -1 \mod 2 \\ x \equiv -1 \mod 3 \\ x \equiv -1 \mod 4 \\ x \equiv -1 \mod 5 \\ x \equiv -1 \mod 6$$
And by the Chinese Remainder Theorem, the solution to the latter is easily found to be $x \equiv -1 \mod{\mathrm{lcm}(2,3,4,5,6)}$, or $x \equiv -1 \mod 60$.
So just count the number of $3$-digit numbers $x$ satisfying $x \equiv -1 \mod 60$. 
A: Exercises: 
Try dividing $420 - 1$ by integers $2, 3, ..., 7$ and note the respective remainders are $1, 2, 3, 4, 5, 6$. 
Then note that $420 = \textrm{lcm}\,(2, 3, 4, 5, 6, 7)$
Any $n = k\cdot 420 - 1$, $k\in \mathbb{Z}$, will yield the same remainders when divided by $2, 3, ..., 7$.

Try dividing $840 - 1 = 839\,$ by $\,2,\, 3, \,4, \,5, \,6, \,7,\,8\,$ $\implies$ respective remainders of $\;$...?...
Then note that $840 = \textrm{lcm}\,(2, 3, 4, 5, 6, 7, 8)$.
So the same respective remainders are obtained, when $k\cdot 840 - 1$ is divided by each of $\{2, 3, ..., 8\}$. 

Now: What\s the least positive number, that when divided by each of $\{2, 3, 4, 5, 6, 7, 8, 9\}$ gives a corresponding remainder of $\{1, 2, 3, 4, 5, 6, 7, 8\}$? 
Find the $\text{lcm}\,(2, 3, 4, 5, 6, 7, 8, 9\} = 2^3\cdot 3^2\cdot 5 \cdot 7 = 840\cdot 3 = 2520.\;$ 
Subtract $1:\;\;$ $2520 - 1 = 2519 = n_9$.
Test by dividing $n_9$ by each of $2, 3, 4, 5, 6, 7, 8, 9$, to confirm...
There you go.
A: You seek numbers which, when divided by $k$ (for $k=2,3,4,5,6$) gives a remainder of $k-1$. Thus the numbers you seek are precisely those which are one less than a multiple of $k$ for each of these values of $k$. To find all such numbers, consider the lowest common multiple of $2$, $3$, $4$, $5$ and $6$, and count how many multiples of this have three digits.
A: Observe that $119$ leaves remainder $n-1$ when divided by $n$  where $2\le n\le 6$
So, $119+1$ is divisible by $n$ for $2\le n\le 6$
So, we numbers of the form $\operatorname{lcm}(2,3,4,5,6)m-1=60m-1$ where $m$ is any positive integer. 
As the required number is of three digits $100\le 60m-1\le 999\implies 2\le m\le 16$ 
Hence there are $16-2+1=15$ such numbers.
A: A brute force solution in python:
    >>> def check_remainders(three_digit_number):
    ...     return all(three_digit_number % n == n -1
    ...         for n in (2, 3, 4, 5, 6))
    ...
    >>> # Filter a list of all 3-digit numbers with the above function
    ... all_numbers = filter(check_remainders, range(100, 1000))
    >>> print(all_numbers)
    [119, 179, 239, 299, 359, 419, 479, 539, 599, 659, 719, 779, 839, 899, 959]
    >>>
    >>> # "Try adding multiples of 60 to 119", Gerry Myerson said.
    ... [119 + (i * 60) for i in range(15)]
    [119, 179, 239, 299, 359, 419, 479, 539, 599, 659, 719, 779, 839, 899, 959]
    >>>
    >>> # so how many numbers like these are there?
    ... print(len(all_numbers))
    15

