Counting for ages

I have four children.

  • This year, the age of the oldest one is equal to the product of the ages of his siblings.
  • Next year, his age will be equal to the sum of their ages.
  • How old is the oldest one?

Note: All children have different ages (i.e. there are no twins).


The answer is 10.

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4 solutions

Calvin Lin Staff
Oct 16, 2020

WLOG let a > b > c > d a > b > c > d .
We have a = b c d a = bcd and a = b + c + d + 2 a = b+c+d + 2 .

Since b c d = b + ( c + 1 ) + ( d + 1 ) < b + b + b = 3 b bcd = b + (c+1) + (d+1) < b + b + b = 3b , so c d < 3 cd < 3 which means c d = 2 , c = 2 , d = 1 cd=2, c = 2, d = 1 .
From there, solving 2 b = b + 2 + 1 + 2 2b = b + 2 + 1 + 2 gives us b = 5 b = 5 .
Thus a = b c d = 10 a = bcd = 10 .

How is d=2,c=1 where c>d? I think it should be c=2,d=1.

Sudipto Podder - 7 months, 3 weeks ago

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Fixed thanks.

Calvin Lin Staff - 7 months, 3 weeks ago

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

Gazar Khalid - 7 months, 3 weeks ago

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Since b > c > d b > c > d , the difference between b b and d d at least has to be 2. So b + ( c + 1 ) + ( d + 1 ) b + (c+1) + (d+1) at most is 3 b 1 3b - 1

Valentino Wu - 7 months, 3 weeks ago
Toby M
Oct 16, 2020

Too long for a comment:

Set up the equations a = b c d a = bcd and a = b + c + d + 2 a = b + c + d + 2 , again with a > b > c > d a > b > c > d without loss of generality.

Observe that for large values of b , c , d b,c,d , b c d bcd is much greater than b + c + d + 2 b+c+d+2 . So if a a is odd, all three of b c d bcd must be odd. The first instance where this happens is when b = 5 , c = 3 , d = 1 b=5, c=3, d=1 , but 5 3 1 > 5 + 3 + 1 + 2 5 \cdot 3 \cdot 1 > 5 + 3 + 1 + 2 . So a a must be even. Similarly, b , c , d b,c,d cannot all be even as 6 4 2 > 6 + 4 + 2 + 2 6 \cdot 4 \cdot 2 > 6 + 4 + 2 + 2 .

So in b , c , d 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 a = b + c + d + 2 implies that a 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 a=b\times c \times d and a = b + c + d + 2 a = b+c+d+2

smallest possible solution is a = 10 , b = 5 , c = 2 a = 10,b = 5,c=2 and d = 1 d =1

How did you find that solution? Are there any other solutions?

Calvin Lin Staff - 3 years, 5 months ago

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Hit and Trial are always helpful. this is what I found the Smallest.

Rishabh Deep Singh - 3 years, 5 months ago

Guys is there a proper solution to this question , other than hit and trial?

Gazar Khalid - 7 months, 3 weeks ago

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I've added a proper solution.

Calvin Lin Staff - 7 months, 3 weeks ago

This is actually the same as Calvin Lin's solution but I'm trying to explain it more thoroughly.

Assume d < c < b < a d<c<b<a with each being a natural number. From the question we have a = b c d a=bcd and a = b + c + d + 2 a=b+c+d+2 , and so b c d = b + c + d + 2 bcd=b+c+d+2 .

Since a , b , c , d a,b,c,d are all natural numbers, from the assumption we get:

c < b c + 1 b c<b\implies c+1\leq b . Hence, d < c < b d + 2 c + 1 b d<c<b \implies d+2\leq c+1\leq b .

Since c < b c<b and d + 2 b d+2\leq b , we have b c d = b + c + ( d + 2 ) < 3 b bcd=b+c+(d+2)<3b , so c d < 3 cd<3 that gives the only possible solution being d = 1 , c = 2 d=1,c=2 .

Insert them to the equation above we get 2 b = b + 5 2b=b+5 so b = 5 b=5 . Hence, a = b c d = 10 a=bcd=10 .

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