# Powers of two my children be

Powers of two my children be,
Arriving in birth years separately.

Their sum is now the reverse of me
And the years between me and a power of three.

My digit difference is the number you see
And my digits are powers of binary.

And now the question I ask of thee:
Their number and age, and the age that I be.

A little riddle I came up with for my sons birthday.

• i hope your son is a genius – pietz Mar 28 '18 at 7:11
• @pietz: Well, I didn't make it for him. I just came up with it on his birthday, since he was the last one to turn a power of two. – gnovice Mar 29 '18 at 14:28

Powers of two my children be,
Arriving in birth years separately.


Unspecified number of children but no twins and their age is 2^n.

Their sum is now the reverse of me


Their sum can be any number since any number can be expressed as a sum of powers of two. Your age can be written 10d + u and the sum of their age 10u + d.

And the years between me and a power of three.


10d + u - 3^x = 10u + d or 3^x - (10d + u) = 10u + d
Therefore 3^x = 9(u - d) or 3^x = 11(u + d)
The latter has no solution but the former has many solutions: (u - d) just needs to be a power of three

My digit difference is the number you see


The difference is the number of children: you have either three or nine children.

And my digits are powers of binary.


There are four possible digits: 1, 2, 4 and 8
If their difference is a power of three (= 3), they can only be 1 and 4.

And now the question I ask of thee:
Their number and age, and the age that I be.


You are 41 and the sum of their age is 14 (1110 in binary). Therefore you have three children and they are 8, 4 and 2 years old.

• You got it! And very nice proof by the way. I wasn't even expecting a formula since I actually worked out the clues to give by just brute-force checking all combinations. – gnovice Mar 27 '18 at 16:33
• Thank you @gnovice That was a nice, fun-to-solve puzzle. – xhienne Mar 27 '18 at 16:37
• @xhienne Where do you get the 10 in 10d+u from? – Orphevs Mar 27 '18 at 21:40
• @Orphevs: If my age is made of two individual digits d and u, then my age will be 10*d+u (i.e. d is in the tens place, u is in the ones place). – gnovice Mar 27 '18 at 21:47
• @gnovice Thank you, I forgot about base 10 for a second. – Orphevs Mar 27 '18 at 21:49

I know this answer is pretty unlikely, but it DOES fit the hints. You are 21, kids are 8 and 4.

Powers of two my children be,

8 and 4 are 2^3 and 2^2

Arriving in birth years separately.

Not the same age!

Their sum is now the reverse of me

8 + 4 = 12 which is the reverse of 21

And the years between me and a power of three.

21 - 9 = 12. 9 is a power of 3, 3^2

My digit difference is the number you see

I don't know what this means

And my digits are powers of binary.

2 = 2^1. 1 = 2^0

And now the question I ask of thee:

Their number and age, and the age that I be.

You are 21, kids are 8 and 4.

• So, I had a kid at 13, eh? ;) We're getting closer! It satisfies all the criteria except the fifth line. – gnovice Mar 27 '18 at 16:15
• It seemed unlikely but you never know! – David Foong Mar 27 '18 at 16:16

Your children are 32 and 4. You are 63.

Powers of two my children be,

Arriving in birth years separately.

4 and 32 are powers of 2. They'd obviously be born in separate years.

Their sum is now the reverse of me

32+4=36. 63 Is the reverse.

And the years between me and a power of three.

63-36=27, a power of 3. (36 being their sum and 63 being your age.)

And my digits are powers of binary.

Least sure about this one - not sure what a "power of binary" really means. But at a guess, 63 is 0b111111 - all the digits are 1.

• How about the fifth line? And "powers of binary" was just another way of saying "powers of two" in a way that rhymed. ;) Also, you should try to use spoiler text. – gnovice Mar 27 '18 at 15:53
• @gnovice Ah, I missed that line! And I figured it may not be right considering the last line anyway, but thought I'd give it a go. – berry120 Mar 27 '18 at 16:56