This Can Drive You Nuts!

Algebra Level 3

The 3 Wilson brothers and the 2 Martin brothers are sharing a bag of peanuts.
Ash Martin eats one more than a quarter of the whole lot.
Bill Wilson then eats one more than a quarter of the remainder.
Chip Martin eats one more than a quarter of what was left.
And Doug Wilson eats one more than a quarter of the remainder.
At this stage the Martin boys had eaten exactly 100 peanuts more than the Wilson brothers.
Finally, Earl Wilson finishes off all the remaining peanuts that were left.

The 3 Wilson boys ate more peanuts than the 2 Martin brothers, but exactly how many more?

This problem is not an original. It is adapted from a problem posed by Sam Loyd.


The answer is 220.

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2 solutions

Satyen Nabar
Mar 8, 2014

Let P be the number of peanuts.

AM (Ash Martin) gets 1+P/4 = (4+P)/4.

Remainder is P - (4+P)/4 = (3P-4)/4

BW (Bill Wilson) gets 1+ 1/4{ (3P-4)/4} = (12+3P)/16

Remainder is (3P-4)/4 - (12 +3P)/16= (9P-28)/16

CM (Chip Martin) gets 1+ 1/4(9P-28)/16= (36+9P)/64

Remainder is (9P-28)/16 - (36+9P)/64= (27P-148)/64

DW (Doug Wilson) gets 1+1/4(27P-148)/64= (27P+108)/256

AM+ CM- BW- DW=100

Solving that you get P= 1020.

AM gets 256

BW gets 192

CM gets 144

DW gets 108

And EW ( Earl Wilson) gets 320.

The 3 Wilson brothers ate 220 peanuts more than the 2 Martin boys

yeap , i also got it right

Mamberry Touray - 7 years, 2 months ago

same method..

Rishabh Tripathi - 7 years, 2 months ago

I was hoping there would be a better metod in the comments...

Luis Filipe Martins Barros - 7 years, 2 months ago

I gave up cause I thought there was a better method...too...

Julian Poon - 7 years, 2 months ago

Log in to reply

Me too....... :P

Harsh Shrivastava - 7 years ago

finally i got it.........

Swapnil Pardeshi - 7 years, 2 months ago

wow! i never thought that i would have to go through all that method and solutions to get the right answer. lol. Well Done!

collins emurerhi - 7 years, 2 months ago

Got it ryt!!!

avilash mahapatra - 7 years, 2 months ago

This hardly qualifies as a Level 3 problem.

Aronas Nuresi - 7 years, 1 month ago
Sung Cho
Apr 12, 2014

Let S be the number of total peanuts.

A gets 1 4 S + 1 = 1 4 ( S + 4 ) \frac{1}{4}S+1 = \frac { 1 }{ 4 } (S+4) .

B gets 1 4 ( S 1 4 ( S + 4 ) ) + 1 = 3 16 ( S + 4 ) \frac { 1 }{ 4 } (S-\frac { 1 }{ 4 } (S+4))+1 = \frac { 3 }{ 16 } (S+4)

C gets 1 4 ( S 3 16 ( S + 4 ) ) + 1 = 9 64 ( S + 4 ) \frac { 1 }{ 4 } (S-\frac{3}{16}(S+4))+1 = \frac{9}{64}(S+4)

D gets 1 4 ( S 9 64 ( S + 4 ) ) + 1 = 27 256 ( S + 4 ) \frac{1}{4}(S-\frac{9}{64}(S+4))+1 = \frac{27}{256}(S+4)

(See the rule here? n t h { n }^{ th } person gets 3 ( n 1 ) 4 n ( S + 4 ) \frac { { 3 }^{ (n-1) } }{ { 4 }^{ n } } (S+4) number of peanuts.)

Since A+C=B+D+100,

( S + 4 ) ( 1 4 + 9 64 ) = ( S + 4 ) ( 3 16 + 27 256 ) + 100 (S+4)(\frac{1}{4} + \frac{9}{64}) = (S+4)(\frac{3}{16}+\frac{27}{256}) + 100

Which is, 25 256 ( S + 4 ) = 100 \frac{25}{256}(S+4)=100

Therefore S = 1020.

Now we know that A=256, B=192, C=144, D=108, and E=320 (E got the remainder)

The difference is (B+D+E)-(A+C), which is 220 peanuts,

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