22
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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.

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  • 1
    $\begingroup$ i hope your son is a genius $\endgroup$ – pietz Mar 28 '18 at 7:11
  • $\begingroup$ @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. $\endgroup$ – gnovice Mar 29 '18 at 14:28
13
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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.

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  • $\begingroup$ 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. $\endgroup$ – gnovice Mar 27 '18 at 16:33
  • $\begingroup$ Thank you @gnovice That was a nice, fun-to-solve puzzle. $\endgroup$ – xhienne Mar 27 '18 at 16:37
  • $\begingroup$ @xhienne Where do you get the 10 in 10d+u from? $\endgroup$ – Orphevs Mar 27 '18 at 21:40
  • 1
    $\begingroup$ @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). $\endgroup$ – gnovice Mar 27 '18 at 21:47
  • $\begingroup$ @gnovice Thank you, I forgot about base 10 for a second. $\endgroup$ – Orphevs Mar 27 '18 at 21:49
3
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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.

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  • $\begingroup$ So, I had a kid at 13, eh? ;) We're getting closer! It satisfies all the criteria except the fifth line. $\endgroup$ – gnovice Mar 27 '18 at 16:15
  • 3
    $\begingroup$ It seemed unlikely but you never know! $\endgroup$ – David Foong Mar 27 '18 at 16:16
1
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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.

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  • 3
    $\begingroup$ 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. $\endgroup$ – gnovice Mar 27 '18 at 15:53
  • $\begingroup$ @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. $\endgroup$ – berry120 Mar 27 '18 at 16:56

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