Makes me wonder
If a , b , c , d are real values that satisfy the system of equations:
a
b
c
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a
b
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b
c
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c
a
+
a
+
b
+
c
=
7
1
b
c
d
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b
c
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c
d
+
d
b
+
b
+
c
+
d
=
1
9
1
c
d
a
+
c
d
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d
a
+
a
c
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c
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d
+
a
=
9
5
d
a
b
+
d
a
+
a
b
+
b
d
+
d
+
a
+
b
=
1
4
3
Evaluate the value of a b c d + a + b + c + d .
The answer is 227.
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3 solutions
correct
A 13 year old wondering about such thing....it happens only in Asia
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Where that comment came from? regardless ...
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I meant that at the age of 13 i used to wonder about cricket not algebra and the guy posting this is 13
Wrote a C program for it . LOL :p
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Can you give me the code?I am interested.
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If you want to write a code for this problems, you can find it in the repository ..
Can you please explain the part where you used the LCM to find out each term's value?
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L e t s c o n s i d e r a n a r b i t r a r y p r o b l e m 2 a = 3 b = 4 c , f o r p o s i t i v e i n t e g e r s a , b a n d c . A s s u m e t h a t t h e s e e q u a l i t y e q u a l s s o N ⇒ 2 a = 3 b = 4 c = N ⇒ a = 2 N , b = 3 N , c = 4 N A s a i s i n t e g e r s o N m u s t b e d i v i s i b l e b y 2 . A s b i s i n t e g e r s o N m u s t b e d i v i s i b l e b y 3 . A s c i s i n t e g e r s o N m u s t b e d i v i s i b l e b y 4 . S o y o u c a n s i m p l y g o f o r N = 2 × 3 × 4 = 2 4 , b u t w h e n w e t a l k a b o u t s m a l l e s t p o s s i b l e v a l u e o f N , w e m u s t g o f o r L C M . 2 = 1 × 2 3 = 1 × 3 4 = 1 × 2 × 2 S o , N = L C M ( 2 , 3 , 4 ) = 1 × 2 × 2 × 3 = 1 2 . O b v i o u s l y L C M ( a , b , c ) i s d i v i s i b l e b y a , b a n d c . N o w a = 6 , b = 4 , c = 3 I n g e n e r a l a = 6 k , b = 4 k , c = 3 k , f o r k = 1 , 2 , 3 , . . . .
wOw amazing....
really solved it like that .......but really nice problem..:D
Nice work. Thanks.
Scholar !!
Factor each equation to (a+1)(b+1)(c+1)=72 etc. Multiply them all and take the cube root to get (a+1)(b+1)(c+1)(d+1)=576. Now you can use this and your previous equations to find a+1 etc. and thus a, you will get a=2 b=5 c=3 d=7 so the answer is 227.
Solved the same way...
Nice work.
so good!!!
Adding 1 in each equation we get
(a+1)(b+1)(c+1)=72,
(b+1)(c+1)(d+1)=192,
(c+1)(d+1)(a+1)=96,
(a+1)(b+1)(d+1)=144,
Now multiply All the equations and take cube on both sides.
Then we get,
(a+1)(b+1)(c+1)(d+1)=(72 X 192 X 96 X 144)^(1/3)=576,
Now divide this by each of the earlier equations to get
d+1=8, a+1=3, b+1=6, c+1=4,
i.e, a=2 , b=5, c=3, d=7,
So (abcd+a+b+c+d)=227
Nice work.
any other way
so great!!
We can factorize the equations-
( a + 1 ) ( b + 1 ) ( c + 1 ) = 7 2 ( b + 1 ) ( c + 1 ) ( d + 1 ) = 1 9 2 ( c + 1 ) ( d + 1 ) ( a + 1 ) = 9 6 ( d + 1 ) ( a + 1 ) ( b + 1 ) = 1 4 4
And equations can be reformed as-
( a + 1 ) ( b + 1 ) ( c + 1 ) ( d + 1 ) = 7 2 ( d + 1 ) ( b + 1 ) ( c + 1 ) ( d + 1 ) ( a + 1 ) = 1 9 2 ( a + 1 ) ( c + 1 ) ( d + 1 ) ( a + 1 ) ( b + 1 ) = 9 6 ( b + 1 ) ( d + 1 ) ( a + 1 ) ( b + 1 ) ( c + 1 ) = 1 4 4 ( c + 1 )
So, we get, 1 9 2 ( a + 1 ) = 9 6 ( b + 1 ) = 1 4 4 ( c + 1 ) = 7 2 ( d + 1 )
⇒ 8 ( a + 1 ) = 4 ( b + 1 ) = 6 ( c + 1 ) = 3 ( d + 1 )
As we know the Lowest Common Multiple of (8, 4, 6, 3) = 24
So, 8 ( a + 1 ) = 2 4 ⇒ a = 2 4 ( b + 1 ) = 2 4 ⇒ b = 5 6 ( c + 1 ) = 2 4 ⇒ c = 3 3 ( d + 1 ) = 2 4 ⇒ d = 7
Thus a b c d + a + b + c + d = 2 2 7