Piles of matchsticks

Algebra Level 3

Gardner emptied his box of matches on the table, distributing them in three piles with different amounts of matches. Among them there are a total of 48 matches.

Then he proposed the following puzzle:

If the first pile goes to the second as many matches as there are in it, after the second pile I move to the third as many matches as there are in the third, and finally, from the third pile, the first match is as many as there are now in the first heap, it turns out that there will be the same number of matches in each pile. How many matches were in each pile?

(22; 14; 12) (20, 16, 12) (8; 16; 24) (8; 28; 12)

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1 solution

Francis Naldo
May 17, 2018

Let a , b , c a, b, c be the original number of matches on each pile and A , B , C A, B, C be the current number of matches on each pile.

First step-

Take B B matches from the first pile

A = A B A = a b A = A - B \implies A = a - b

then move them to the second pile

B = B + B 2 × b B = B + B \implies 2 \times b .

Second step-

Take C C matches from the second pile

B = B C B = 2 × b c B = B - C \implies B = 2 \times b - c

then move them to the third pile

C = C + C C = 2 × c C = C + C \implies C = 2 \times c .

Third step-

Take A A matches from the third pile

C = C A C = 2 × c ( a b ) C = C - A \implies C = 2 \times c - (a - b)

then take them to the first pile

A = A + A A = 2 × ( a b ) A = A + A \implies A = 2 \times (a - b) .

Next step-

The total number of matches is 48 48

a + b + c = 48 a + b + c = 48 and A + B + C = 48 A + B + C = 48

and since A = B = C A = B = C

then,

A = B = C = 48 3 = 16 A = B = C = \frac{48}{3} = 16 .

Last step-

Simply the three equations.

[ 2 2 0 0 2 1 1 1 2 ] [ a b c ] = [ 16 16 16 ] \begin{bmatrix} 2 & -2 & 0\\ 0 & 2 & -1\\ -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix} = \begin{bmatrix} 16\\ 16\\ 16 \end{bmatrix}

Three equations for three variables. lots of possible way to solve from this point but here's one way.

From eqn1,

2 a 2 b = 16 a = 8 + b 2a - 2b = 16 \implies a = 8 + b

from eqn2,

2 b c = 16 c = 2 b 16 2b - c = 16 \implies c = 2b - 16

substitute to eqn3,

a + b + 2 c = 16 ( 8 + b ) + b + 2 ( 2 b 16 ) = 16 8 b + b + 4 b 32 = 16 b = 14 -a + b + 2c = 16 \implies -(8 + b) + b + 2(2b - 16) = 16 \implies -8 - b + b + 4b - 32 = 16 \implies \fbox{b = 14}

using this, get a a and c c by substituting (b) to eqn1 and eqn2,

a = 8 + b a = 8 + ( 14 ) a = 22 a = 8 + b \implies a = 8 + (14) \implies \fbox{a = 22}

c = 2 b 16 c = 2 ( 14 ) 16 c = 12 c = 2b - 16 \implies c = 2(14) - 16 \implies \fbox{c = 12} .

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