JUST A MODULO PROBLEM!!

since, we have to find x,

we will proceed by using Remainder theorem or the Modulo method!

given,

x=9(mod10)

x=8(mod9)

x=7(mod8)

x=6(mod7)

x=5(mod6)

x=4(mod5)

x=3(mod4)

x=2(mod3)

x=1(mod2)

thus, adding 1 on both sides we get,

x+1=0(mod10)

x+1=0(mod9)

x+1=0(mod8)

x+1=0(mod7)

x+1=0(mod6)

x+1=0(mod5)

x+1=0(mod4)

x+1=0(mod3)

x+1=0(mod2)

thus, (x+1) is divisible by 10,9,8.....2.

thus, it the L.C.M. of 10,9,8.....2.

=(10)x(9)x(4)x(7)

=2520 thus, the answer is x, one less than 2520

=2519

Nice solutions, I went for the code :)

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c=0
while True:
    works=True
    a=9
    while a>2 and works:
        if c%a==a-1:
            works=True
        else:
            works=False
        a-=1
    if works:
        print c
        break
    c+=1

lcm of the integers -1