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Tuesday puzzle, An easy one to start with ;-)

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Tuesday puzzle, An easy one to start with ;-)

#1

Tuesday puzzle, An easy one to start with ;-)

I have a 5 digit number A B C D E with each digit being from 1 to 9 and no duplicates. Two are prime, two are squares and one is neither. The 3rd digit is twice the 5th, the 4th digit is 6 more than the second digit and the first digit is 3 less than the 5th. What is the number?

Re: Tuesday puzzle, An easy one to start with ;-)

#2

Re: Tuesday puzzle, An easy one to start with ;-)

I'm not sure how to answer this without going to the box since it is early in the day.....I believe the unused numbers are 2, 5, 6 & 7.

There is a WIFI hot spot at 830 W Ave P Palmdale, CA with the final number.

Lee

Re: Tuesday puzzle, An easy one to start with ;-)

#3

Re: Tuesday puzzle, An easy one to start with ;-)

"I believe the unused numbers are 2, 5, 6 & 7. "

But that would imply that 1, 4, and 9, all THREE of which are perfect squares, are in the number. The problem states that there are two squares (Which I interpreted to mean "exactly two squares".

Re: Tuesday puzzle, An easy one to start with ;-)

#4

Impossible?

"The 3rd digit is twice the 5th"

"the first digit is 3 less than the 5th"

If the first digit is n, this says that the 3rd digit is 2*(n+3), which is a digit only when n is 0 (not allowed) or 1. So that gets us to

1x8y4

accounting for the two squares and the one "neither". So x and y are our primes. There are not two prime digits that satisfy

"the 4th digit is 6 more than the second digit"

Where have I gone wrong?

Re: Tuesday puzzle, An easy one to start with ;-)

#5

Creative definitions?

If you considered 1 a prime (it's not), and

if you considered 1 not a perfect square (it is), then

13894

would be a solution, consistent with Lee's hint of his solution.

Re: Tuesday puzzle, An easy one to start with ;-)

#6

Re: Creative definitions?

I also came up with 13894 but it was too early in the day to spoil the fun.

If the argument that 1 is a square of itself, wouldn't it also be a prime number since a prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself.? It's been too many years away from math classes in school to remember all the definitions.

Lee

Re: Tuesday puzzle, An easy one to start with ;-)

#7

Re: Creative definitions?

"If the argument that 1 is a square of itself, wouldn't it also be a prime number since a prime number is a natural number which has exactly two **distinct** natural number divisors: 1 and itself.?"

"Distinct" is important here. 1 and 1 are not distinct, or if they were, we could say that one has three distinct divisors, 1, 1, and 1, and is thus not a prime.

According to the Wikipedia article, before the 19th century, one was considered a prime, but no more.

Re: Tuesday puzzle, An easy one to start with ;-)

#8

oops

mistake, should be last is 3 less than first. That is what I get from typing this stuff so late, sorry.

Re: Tuesday puzzle, An easy one to start with ;-)

#9

Better ;-)

That's a whole different kettle of fish!

Figured it had to be something like that.

Solved quickly this time.

To not spoil the fun, I'll write it as

SQRT(5460323236)

Re: Tuesday puzzle, An easy one to start with ;-)

#10

Re: Better ;-)

See, it really is easy once I post it correctly ;-)

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