A number when divided by 7, 11,and13[the prime factor of 1001] successively leaves the remainder 6, 10,and 12?

Answer 1

1000.
good question but not as good as my mathematics.

Let number=x
Let Qy(x) is a function which returns the quotient when x is divided by y
i.e. Qy(x)=int[x/y] -->where int[] represents the integer part of x
Now,
x=7* Q7(x)+6...................eq1
let Q7(x)=y
y=11* Q11(y)+11.....................eq2
let Q11(y)=z
z=13* Q13(z)+12..........................eq3

by substituing eq3 into eq2 and then the new equation into eq1:
x=7 (11 ( 13 * Q13(z) + 12 ) + 10 ) + 6
x=77 ( 13 * Q13 (z) + 12 ) + 76
x=1001 * Q13 (z) + 924 + 76
x=1001 * Q13(z) + 1000
Now,
Q13(z) = int [ z /13 ] = int [ int [ y /11] /13 ] = int [ int [ int [ x / 7 ] /11 ] / 13 ] = int[ int [ int [ x / 1001] ] ]
Let A=int[ int [ int [ x / 1001] ] ]
Hence,the final equation is :
x=A + 1000
clearly,
A is zero for all values of x<1001 or x<=1000
substituting A=0
x=1000

other cases are also there like:
i) A is 1 if x=1001.....but x can't be 1001, bcoz then remainder will always be 0.
ii)A is 2 if x>=2002.....but then RHS=2+1000=1002 & LHS=2002
clearly,contradicting.
iii) For any other value ,similarly it is contradicting.


Only, possible ans is x=1000

good luck!

Answer 2

1000

7 X 11 X 13 = 1001

6 = 7 - 1
10 = 11 -1
12 = 13 -1
i.e. 1 short of 1001 (the LCM)

1001 - 1 = 1000 will make the above statement true.

Answer 3

That means if we add 1 it is divisible by 7 ,11,13 so it is one less than LCM so the number is 1001n-1 for n >0

Answer 4

1000

Answer 5

42

Answer 6

let num be x
x=7n+6;
x=11m+10;
x=13k+12;
solve for x;
it comes out to be 1000

Answer 7

1000
you are talking here about SUCCESSIVE division , but still 1000 is correct.