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0 / 60 seg.
A farmer has 30 animals consisting of chickens and cows. The animals have 74 legs in total. Find out the number of chickens and cows the farmer has
2
3
C
H
_
C
K
_
N
S
_
N
D
7
C
_
W
S
.
L
_
T
T
H
_
N
_
M
B
_
R
_
F
C
H
_
C
K
_
N
B
_
X
_
N
D
N
_
M
B
_
R
_
F
C
_
W
S
B
_
Y
.
C
H
_
C
K
_
N
S
H
_
V
_
2
_
N
D
C
_
W
S
H
_
V
_
4
L
_
G
S
.
B
_
S
_
D
_
N
T
H
_
S
,
W
_
C
_
N
C
_
M
_
_
P
W
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T
H
2
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Q
_
_
T
_
_
N
S
:
2
X
+
4
Y
=
7
4
X
+
Y
=
3
0
S
_
L
V
_
N
G
F
_
R
X
_
N
D
Y
W
_
L
L
G
_
V
_
2
3
_
N
D
7
R
_
S
P
_
C
T
_
V
_
L
Y
Clue
THERE ARE MANY POSSIBLE SOLUTIONS ONCE YOU GET THE IDEA. BELOW IS ONE POSSIBLE SOLUTION
23 CHICKENS AND 7 COWS. LET THE NUMBER OF CHICKEN BE X AND NUMBER OF COWS BE Y. CHICKENS HAVE 2 AND COWS HAVE 4 LEGS. BASED ON THIS, WE CAN COME UP WITH 2 EQUATIONS: 2X + 4Y = 74 X + Y = 30 SOLVING FOR X AND Y WILL GIVE 23 AND 7 RESPECTIVELY
THE TOTAL IS 2. WHEN THE BOOK ON THE LEFT IS UPSIDE DOWN THEN THE PAGE NUMBER ON THE EXTREME LEFT IS 1 AND THE RIGHT HAND SIDE BOOK'S EXTREME RIGHT HAND SIDE PAGE NUMBER IS ALSO 1. ANOTHER WAY TO LOOK AT THE PROBLEM: IF TWO BOOKS ARE STANDING SIDE BY SIDE TO EACH OTHER, PAGE 1 IS ON THE RIGHT SIDE OF EACH BOOK (ASSUMING YOU ARE LOOKING AT THE SPINE OF THE BOOKS). TURNING THE ONE ON THE LEFT UPSIDE DOWN (WITH THE BOOK SPINE STILL POINTING OUT), BRINGS PAGE 1 TO THE LEFT SIDE. AND SO AT THE "EXTREME LEFT" OF THE BOOK ON THE LEFT IS PAGE 1, WHILE THE "EXTREME RIGHT" OF THE BOOK ON THE RIGHT IS ALSO PAGE 1
142857. THE NUMBER AND THE SUMS WILL ONLY CONTAIN THE DIGITS 1,2,4,5,7 AND 8. 142857 + 142857 = 285714 142857 + 142857 + 142857 = 428571 142857 + 142857 + 142857 + 142857 = 571428 142857 + 142857 + 142857 + 142857 + 142857 = 714285 142857 + 142857 + 142857 + 142857 + 142857 + 142857 = 857142 142857 + 142857 + 142857 + 142857 + 142857 + 142857 + 142857 = 999999
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