X=n-1 (mod n) ((n=2,..,10))

By Nikos Xasiotis on Thursday, October 11, 2001 - 06:34 pm:

Which number gives:
X mod 10=9
X mod 9=8
X mod 8=7
X mod 7=6
X mod 6=5
X mod 5=4
X mod 4=3
X mod 3=2
X mod 2=1

By Kerwin Hui on Thursday, October 11, 2001 - 08:44 pm:

First you get rid of the redundant conditions and you get

$X \equiv 8 (\mod 9)$

$X \equiv 7 (\mod 8)$

$X \equiv 6 (\mod 7)$

$X \equiv 4 (\mod 5)$

which you can now solve by Chinese Remainder Theorem, as the moduli are pairwise coprime.

Another approach is to note that the condition given is the same as

$X \equiv - 1 (\mod n)$

where $n$ ranges from 2 to 10. Now apply a variation of Chinese Remainder Theorem to get the answer.

Kerwin

By Michael Doré on Thursday, October 11, 2001 - 09:53 pm:

Or just note that X + 1 is divisible by 2,3,...,10. So it suffices for X + 1 to be divisible by just 5,7,8,9 (since then X + 1 is automatically divisible by 2,3,4,6). So X + 1 is some multiple of 5x7x8x9.

By Yatir Halevi on Saturday, November 17, 2001 - 11:12 am:

Consider the number 10! (factorial of 10)
all the number from 1 to 10 divide 10!
Now consider the number (10!-1), This number will give a remainder of (d-1) when divided by d, where d is any number from 2 to 10.