I have three children. One is the same age as the first number in my age, another is the same age as the second number in my age, and the third is the same age as the sum of the two numbers in my age. None of the children are the same age and the total of our ages is 45. How old am I?
March 9th, 2011 | Category: 
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The first two children are three year old twins and with their six year old sibling, your kids’ total age adds up to 12 years. The total of all your ages would be 45, since you are 33.
1stchild is 1, 2nd is 4,and the third is 7 , witch is 12 your 33 thats 45
Ok, where to start … Paul, you’re wrong! It clearly states that none of the children are the same age.
Carole, you’re wrong! One child is the first number of my age, the second child is the second number of my age and the third is the sum of the first two.
Therefore, I am not 33 years old.
Keep trying.
1st is 2, 2nd is 7 and 3rd is 9 and you are 27.
You are correct, Carole!
27, the children are 9, 7 and 2.
As my children are different ages, the lowest they could be is 1, 2, 3, and as our ages add to 45 this would make me 39. Working backwards the only answer to make the sum correct is 27 + 2 + 7 + 9 = 45.
Alternatively, we could use algebra and say that I am 10A + B years of age. My children are A, B and A+B years of age. Our ages add to 45, so:
10A + B + A + B + A+B = 45
Collecting like terms together gives:
12A + 3B = 45
Dividing throughout by 3 gives:
4A + B = 15
If A = 1 then B would be 11, which makes no sense for an age. If A = 2 then B = 7 and as all values that work give us a correct answer. So I am 27.
Note how A = 3 would make B = 3, which isn’t allowed. And A > 3 will take us past 15, so there can be no other solutions
You’re today’s winner.