Magic of Math #1
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Evaluate the real value of a b c d + a + b + c + d .
The answer is 227.
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3 solutions
Can you show me how you factroise the eqution by adding 1 both sides.
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= a b c + a b + b c + c a + a + b + c + 1 = a b c + a b + b c + b + c a + a + c + 1 = a b ( c + 1 ) + b ( c + 1 ) + a ( c + 1 ) + 1 ( c + 1 ) = ( c + 1 ) ( a b + b + a + 1 ) = ( a + 1 ) ( b + 1 ) ( c + 1 )
if we take cube roots, we also get ( a + 1 ) ( b + 1 ) ( c + 1 ) ( d + 1 ) = 5 7 6 , 5 7 6 ω , 5 7 6 ω 2
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Need real one.
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not mentioned in question
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@Aareyan Manzoor – @Akshat Sharda , he is right. Please add that to your question.
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@Pi Han Goh – It's not my question.
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@Akshat Sharda – haha LOL... I didn't see it.
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@Pi Han Goh – Still I just edited the question .... please see it now.
Subtract each pAir of equations to get: ( d − a ) ( b c + c + b + 1 ) = ( d − a ) ( b + 1 ) ( c + 1 ) = 5 ∗ 2 4 ( b − a ) ( c d + c + d + 1 ) = ( b − a ) ( c + 1 ) ( d + 1 ) = 4 ∗ 2 4 ( b − c ) ( a d + a + d + 1 ) = ( b − c ) ( a + 1 ) ( d + 1 ) = 2 ∗ 2 4 ( d − c ) ( a b + a + b + 1 ) = ( d − c ) ( a + 1 ) ( b + 1 ) = 3 ∗ 2 4 From these to easily get: a = 2 , c = 3 , d = 7 , b = 5 ⇒ R e s = 2 2 7
or you can "+1" on each equation and you'll see something nice.
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I did exactly that.
I did the same , adding 1 !!
I have done almost as Akshat Sharda. His method is better. However just a little deviation.
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⟹
A
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∴
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5
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3
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7
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∴
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d
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2
2
7
Adding 1 to each equation and then factorizing them would give us , ( a + 1 ) ( b + 1 ) ( c + 1 ) = 7 2 ( b + 1 ) ( c + 1 ) ( d + 1 ) = 1 9 2 ( a + 1 ) ( c + 1 ) ( d + 1 ) = 9 6 ( a + 1 ) ( b + 1 ) ( d + 1 ) = 1 4 4 Multiplying all of them and then taking cube root will give us an equation as follows, ( a + 1 ) ( b + 1 ) ( c + 1 ) ( d + 1 ) = 5 7 6 Now we have to divide the above equation from the equations we obtained be adding 1, Then we will reach to the simple equations as follows, d + 1 = 8 ⇒ d = 7 a + 1 = 3 ⇒ a = 2 b + 1 = 6 ⇒ b = 5 c + 1 = 4 ⇒ c = 3 a b c d + a + b + c + d = 2 1 0 + 2 + 5 + 3 + 7 = 2 2 7