MATHS

A number has 3 digits. When it is divided by 6 or 7, it leaves a remainder of 1. When it is divided by 8 or 11, it leaves 2 remainder of 7.
What is the largest such number?

The answer given is 799

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  1. I suggest you Google something like
    "Chinese Remainder Theorem" and I assume you know about modular arithmetic since I will be using properties of modular arithmetic

    so we have
    z = 1 mod 6
    z = 1 mod 7
    z = 7 mod 8
    z = 7 mod 11
    But, in the Chinese Remainder Therorem we need all the modular arguments to be "relatively prime"
    6 and 8 are not

    BUT, 1 mod 6 = 7 mod 6
    and if 7 mod 6 = 7 mod 8
    then 7 mod 6 = 7 mod 24 , (LCM of 6 and 8 is 24)

    so we can reduce our problem to 3 conditions:
    z = 7 mod 24
    z = 1 mod 7
    z = 7 mod 11
    BUT 7 mod 11 = 7 mod 24
    so we now have

    z = 7 mod 264
    z = 1 mod 7

    starting with the smallest 3 digit number, results of
    7 mod 264 are
    271 539 799 1063 ...

    only 3 cases to check for mod 7
    271 mod 7 = 5
    539 mod 7 = 0
    799 mod 7 = 1 ======> yeahhh, its 799

    NOTE:
    I did not actually have to use the "Chinese Remainder Theorem" since I was fortunate to have 7 mod 24 = 7 mod 11

    If I had ended up with something like
    5 mod 24 and 6 mod 11, it would be much more difficult and I would have to resort to the CRT

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