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avatar+1090 

A frog can either jump 7 cm forward or 11 cm forward; he can never jump backward. If the frog jumps along a number line and starts at 0, what is the largest whole number he cannot possibly jump to in any series of jumps?

 

A. 58   B. 59   C. 60   D. 61   E. NOT

Mathematician  Nov 14, 2014

Best Answer 

 #9
avatar+81154 
+15

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

 

CPhill  Nov 14, 2014
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11+0 Answers

 #1
avatar+91517 
+15

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

Melody  Nov 14, 2014
 #2
avatar+1090 
0

How can I do that without guess and check?

Or is that the only way?

Mathematician  Nov 14, 2014
 #3
avatar+17711 
0

Neat analysis, Melody!

geno3141  Nov 14, 2014
 #4
avatar+91517 
+5

Thanks Gino,  

 

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

Melody  Nov 14, 2014
 #5
avatar+81154 
+15

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

CPhill  Nov 14, 2014
 #6
avatar+1090 
0

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!

Mathematician  Nov 14, 2014
 #7
avatar+91517 
0

Thanks Mathematician :)

Melody  Nov 14, 2014
 #8
avatar+26412 
+15

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

frog1

frog 2

.

Alan  Nov 14, 2014
 #9
avatar+81154 
+15
Best Answer

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

 

CPhill  Nov 14, 2014
 #10
avatar+91517 
0

I really like your method Chris - it is so quick  

Melody  Nov 15, 2014
 #11
avatar+1090 
0

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

Mathematician  Nov 19, 2014

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