If the number is divided by 2,3,4,5,6,7,8,9,10 the remainder is 1,2,3,4,5,6,7,8,9 ;but we can divide by 11.?

Answer 1

So the number is one short of being a multiple of the numbers 2 through 10.
LCM of 2 through 10 is 5*7*8*9 = 2520.

Therefore the number 2519 meets the first condition. It is also a multiple of 11 (229*11 = 2519), so there is your number!

Actually, all numbers of the form (11n+1)*2520 - 1 will do.

Answer 2

Not really sure I understand your question, but:

Some of this information is not presented in the best way possible. Search "Chinese Remainder Theorem", though notice it only applies to relatively prime moduli, so you have to work with a number:

=8 (mod 9)
=7 (mod 8)
=6 (mod 7)
=4 (mod 5)

let x be this number. then x = 9a+8 by the first equivalence. But by the second, 9a=7 (mod 8), so a=8b+7, then x=72b+71. By the third (using the same methodology), 2b=5 (mod 7), so b=6 (mod 7), b=7c+6, x= 504c+503. The last one yields c=5d+4, so x = 2520d+2519. Note that 2519 = 229*11, so depending on d, that's the remainder when x is divided by 11; i.e., x = d (mod 11).

Steve

Answer 3

Divide by 2 leaves a remainder of 1 = the number is odd (then n+1 is even)
Divide by 3 leaves a remainder of 2 = (n+1) is divisible by 3
Divide by 4 leaves a remainder of 3 = (n+1) is divisible by 4

So, n+1 is divisible by 2, 3, 4, 5, 6, 7, 8, 9,10 and n is divisible by 11

(should be easier now)

the smallest number for n+1 is
2*2*2*3*3*5*7 = 2520
and it just so happens that 2519 is divisible by 11.

Answer 4

NUMBER=LCM -difference





LCM:-
2,3,4,5,6,7,8,9,10\2
1,2,3,4,3,7,4,9,5\2
1,2,3,2,3,7,2,9,5\2
1,1,3,1,3,7,1,9,5\3
1,1,1,1,1,7,1,3,5\3
1,1,1,1,1,7,1,1,5\5 therefore LCM=2x2x2x3x3x5x7
1,1,1,1,1,7,1,1\7 = 2520
1,1,1,1,1,1,1
number=LCM -1
=2520 -1
= 2519
Verification:-
2519/11=229

Answer 5

The number is 3628801
3628801/11=329891