hi ,, can u help me solve this math problem ??? we have a stair ,?

Question

if we climb the stairs steps 2 at a time then there will be 1 step remaining ,,, if we climb 3 steps at a time then there will be 2 steps left , if we climb 4 steps at a time then there will be 3 steps left , if we climb 5 steps at a time then there will be 4 steps left , if we climb 6 steps at a time then there will be 5 steps left ,, if we climb 7 steps at a time no steps are left ,,,ok the question is what is the total number of steps and write me the equation or how did u figure it out ,,,, hint ,,, the number of stair steps is 119 ,,, just tell me how u got it ?? pls write the equation ,,,, thank u

Answer 1

2 steps at time=>1 remains, suggests that the number is odd.
3 steps at time=>2 remain, suggests that the sum of its numerals is also of type 3a+2 (for example 5, 8, 11, 14...)
4 steps at time=>3 remain, suggests that the last two numerals are of type 4a+3 (11,15,19, 23,...)
5 steps at time=>4 remain, suggests that the last number is either 4 or 9.
6 steps at time=>5 remain, does not tell much, only that the requested number is of type 6a+5 (or 6a-1)
7 steps at time=> no steps remain means that the number is divisible by 7.
Now do the eliminations and you got the answer.

Answer 2

In fact there are infinitely many solutions. Your conditions give you these congruences:

(1) x = 1 (mod 2)
(2) x = 2 (mod 3)
(3) x = 3 (mod 4)
(4) x = 4 (mod 5)
(5) x = 5 (mod 6)
(6) x = 0 (mod 7).

First, notice that (3) => (1) [if x leaves a remainder of 3 when divided by 4, then x leaves 1 when divided by 2] and (5) => (2) [if x leaves 5 when divided by 6, then x leaves 2 when divided by 3]. This means that our system of congruences may be replaced by the more economical system consisting of (3), (4), (5), and (6).

There is a theorem in Number Theory (called the Chinese Remainder Theorem) which can be used on a linear system of congruences if the moduli are relatively prime in pairs, and of course ours are not. Thus, we would like to replace our system with an equivalent system which does satisfy the required condition. Notice that 4 and 6 are not relatively prime. But we can replace (3) and (5) with a single equivalent congruence, namely x = 11 (mod 12) [if x = 11 (mod 12), then x = 3 (mod 4) and x = 5 (mod 6)].

We now consider the following system:

x = 11 (mod 12)
x = 4 (mod 5)
x = 0 (mod 7).

The CRT tells us that
x = (-1)*5*7*(-1) + (-1)*12*7*4 + 0*12*5*2 (mod 12*5*7) or
x = -301 = 119 (mod 420). Thus, any number of the form
119 + 420n (for an integer n ) is a solution of the given system.