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A no divided by 7 leaves 4 as rem..n when divided by 11 leaves 6 as rem..wt will b remain when it is div by 13...1000<no<2000.

i found 39 as lowest.....here after how to proceed...the no is of nature 77a+39

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A no divided by 7 leaves 4 as rem..n when divided by 11 leaves 6 as rem..wt will b remain when it is div by 13...1000<no<2000.

A = 4 (mod 7)

    = 6 (mod 11)

For A = 4 (mod 7), possible numbers are 11, 18, 25 . .  ..

and the first number in the series divisible by 11 is 11

So N₁ = 11

For A  = 6 (mod 11), possible numbers are 17, 28, 39 . . . .

and the first number in the series divisible by 7 is 28

So N₂ = 28

Thus A = N₁ + N₂ = 39

And all the numbers of the form 39 + 77K satsify the above two conditions.