I have four children.
Note: All children have different ages (i.e. there are no twins).
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How is d=2,c=1 where c>d? I think it should be c=2,d=1.
Sir , how did you assume that b+ (c+1) + (d+1) < 3b ? I know that it leads to the correct result but how do we make that statement ? Like what is the basis of the statement
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Since b > c > d , the difference between b and d at least has to be 2. So b + ( c + 1 ) + ( d + 1 ) at most is 3 b − 1
Too long for a comment:
Set up the equations a = b c d and a = b + c + d + 2 , again with a > b > c > d without loss of generality.
Observe that for large values of b , c , d , b c d is much greater than b + c + d + 2 . So if a is odd, all three of b c d must be odd. The first instance where this happens is when b = 5 , c = 3 , d = 1 , but 5 ⋅ 3 ⋅ 1 > 5 + 3 + 1 + 2 . So a must be even. Similarly, b , c , d cannot all be even as 6 ⋅ 4 ⋅ 2 > 6 + 4 + 2 + 2 .
So in b , c , d , there must be either 1 odd, 2 even, or 1 even, 2 odd. When there is 1 odd number and 2 even numbers, a = b + c + d + 2 implies that a is odd., which is a contradiction. So there must be 1 even and 2 odd.
let a be the Oldest one so we get two equations
a = b × c × d and a = b + c + d + 2
smallest possible solution is a = 1 0 , b = 5 , c = 2 and d = 1
How did you find that solution? Are there any other solutions?
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Hit and Trial are always helpful. this is what I found the Smallest.
Guys is there a proper solution to this question , other than hit and trial?
This is actually the same as Calvin Lin's solution but I'm trying to explain it more thoroughly.
Assume d < c < b < a with each being a natural number. From the question we have a = b c d and a = b + c + d + 2 , and so b c d = b + c + d + 2 .
Since a , b , c , d are all natural numbers, from the assumption we get:
c < b ⟹ c + 1 ≤ b . Hence, d < c < b ⟹ d + 2 ≤ c + 1 ≤ b .
Since c < b and d + 2 ≤ b , we have b c d = b + c + ( d + 2 ) < 3 b , so c d < 3 that gives the only possible solution being d = 1 , c = 2 .
Insert them to the equation above we get 2 b = b + 5 so b = 5 . Hence, a = b c d = 1 0 .
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WLOG let a > b > c > d .
We have a = b c d and a = b + c + d + 2 .
Since b c d = b + ( c + 1 ) + ( d + 1 ) < b + b + b = 3 b , so c d < 3 which means c d = 2 , c = 2 , d = 1 .
From there, solving 2 b = b + 2 + 1 + 2 gives us b = 5 .
Thus a = b c d = 1 0 .