Number adding possibilities

Here's one more final take on this problem for me.....note that no number is divisible by 7 or 11.

This means that, at least one 7 and one 11 must be used to form any possible number. Thus, 18 will be used in the addition process.

So......

61 - 18 = 43 and neither 7 or 11 divides this......43 - 18  = 25  and neither 7 or 11 divides this....25 - 18 = 7 and 7 divides itself

So.....61  =  3(18) + 7 =  3(7 + 11) + 7  = 4(7) + 3(11)

60 - 18 = 42   and note that 7 divides this....so 60 = 1(18) + 6(7) = 1(11 + 7) + 6(7) = 7(7) + 1(11)

58 - 18 = 40   ......40 - 18 = 22......22 - 18 = 4  .....note, however.....58 - 11 = 47  .......47 - 11 = 36 and 18 divides this

So...58 =  2(11) + 2 (18)   = 2(11) + 2(11 + 7) = 4(11) + 2(7)

And we have seen that 59 cannot be made by any linear combo of 7 and 11

7L+11B

umm lets see

7,14,28,35,42,49,56,

can't make any of those with 7*8=56

49+11=60    So I can make 60                                   ANSWER IS NOT 60

42+11=53+11=64  can't make any using  7*6

35+11=46+11=57 can't make any using  7*5

28+11=39+11=50+11=61        SO 4*7+3*11=61     ANSWER IS NOT 61

14+4*11=58       so   2*7+4*11=58                         ANSWER IS NOT 58

7+5*11 = 62

So the answer is 59

How can I do that without guess and check?

Or is that the only way?

Neat analysis, Melody!

Thanks Gino,  

Mathematician, I think it has to be a systematic analysis but getting it organised properly is the tricky bit :)

58 = 4(11) + 2(7)

60 = 1(11) + 7(7)

61 = 3(11) + 4(7)

59 is the answer...to see this.....keep subtracting 7s or 11s from it....no resulting number is divisible by 11 or 7

 Also.....the equation of the line  z = 7x + 11y  would pass through all the values except z = 59

Hm...I'm going to think about it a bit more, and see if I can come up with a more formula-like way. If I do I'll tell you!

Thanks Mathematician :)

Here's my (somewhat long-winded) reasoning:

.

I really like your method Chris - it is so quick  

Yes, it is a good method...the best one here, perhaps. But they are all useful.