# Another number theory problem

• Trail_Builder
In summary, the math quiz asked if N has a prime factorization of the form 2^i*3^j*5^k. Homework statement was that N has a prime factorization of the form 2^i*3^j*5^k. The Attempt at a Solution is that N has a prime factorization of the form 2^i*3^j*5^k. The Attempt at a Solution explained that 10*N=4a'^2 and 5*N=2a'^2. The Attempt at a Solution also explained that 2^{x},5^{y}, k_{1}^{z_{1}}, k_{2}^{z_{2}}... | N where x

#### Trail_Builder

this is question found as part of a 25question 1hour long maths problem quiz i took today. It was just about the only one i couldn't do in the time, and even with another look after I can't do it :S I think it may involve maths I havnt come across or at least methodology i havn't (I 16 and only just started looking beyond class studies)

## Homework Statement

Let N be the smallest integer such that 10 x N is a perfect square and 6 x N is a perfect cube.
How many positive factors does N have?

(options)

A 30 B 40 C 54 D 72 E 96

## The Attempt at a Solution

right, Ill talk you through my attempt so far...

so,

I first of all tried a few values for N (the basics as tests like 0,1,2,3,5,10, etc..) and came to the logical conclusion that it wasn't going to be as simple as that, and probably would be a very large number (probably should have concluded that from the options but o well

o yeh, I found out that 0 kinda works, but I'm guessing from the options that 0 doesn't count because 0 isn't a cube or a square? am I right?

Then I tried using modular arithematic (I still a super noobie with it)

N = 0 (mod 10) is a perfect square, while N = 0 (mod 6) is a perfect cube.

From that I started working my way through the first half, trying lots of options for what makes the first bit work. I came up with stuff like N = 10, 40, 90, etc.. Not in a very logical approach but hey, mighta saved me time, but it didnt...

I can't figure out how to find N?

Have I approached the question all wrong though? is finding N entirely necessary or can you derive the factors somehow?

I tried the later approach, looking for similarities between the factors of squares and cubes but I don't know if it was me being slow, or wierd, or whatever, (or oblvivious to the obvious) but I couldn't find anything that way...

and as per usualy, one step at a time ;)

thnx

Clearly N has a prime factorization of the form 2^i*3^j*5^k. What constraints do your premises put on i,j,k?

:S what's prime factorization?

thnx

you can play some tricks involving divisibility.

for instance, you have 10*N is a perfect square. so
$$10N=a^2$$
so 2 divides both sides, hence 2 divides a^2, so 2 divides a. (if 2 doesn't divide a, then 2 can't divide a^2 because 2 is a prime)

so you have:
$$a^2=(2a')^2$$
or
$$a^2=4a'^2$$

substitute,
$$10N=4a'^2$$

cancel out the two
$$5N=2a'^2$$

rinse and repeat for 5. and similarly for 6N=b^3

Sorry to be off topic but was that test the AMC?

If $$10N = a^2$$, then $$2^{x},5^{y}, k_{1}^{z_{1}}, k_{2}^{z_{2}}... | N$$ where x, y can be expressed as 2*q - 1, where q is any positive integer. k refers to any prime, and z an even number.

For $$6N = b^{3}$$, we have $$2^{a},3^{b}, l_{1}^{c_{1}}... | N$$ where a, b can be expressed as 3*q - 1, where q is any positive integer. l refers to any prime and c is a multiple of 3.

Now we have concluded that 2, 3 and 5 must be factors of N. Also, we know what the factors of the respective exponents must be. Can you go on from there?

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Feldoh said:
Sorry to be off topic but was that test the AMC?

it was the second round of the UKMT Maths Challenge (from the UK), and is called something like "the european pink kangaroo round", haha, funny name, lol

anyways, yeh i reks i didnt through to the olympiad cause i working out i probs only got like 90 outa 135 or whatever it is...

whats the AMC?

o yeh, thnx for the help, ill try and go from there and get back to you if i get stuck...

Trail_Builder said:
o yeh, thnx for the help, ill try and go from there and get back to you if i get stuck...

If you are going to keep looking at number theory problem it would be good to get to know about prime factorization. It's not even hard.

o rite i know what prime factorization is lol

just never heard it called that before cause at school i just get told it when you split a number into prime factors hehe, stupid me

right I've decided i have no idea what's going on lol

unfortunately, i not sure what any of your explanations mean or where to go from there.

thnx for the help anyways, ill come to this problem once i read some number theory

So then do you agree N=2^i*3^j*5^k for some i,j,k? You can do it now.

In order that 10*N be a square, since 10= 2*5, N must have factors of 2 and 5. In order that 6*N be a cube, since 6= 2*3, N must have factors of 22 and 32. Since N is the smallest such number, it must be 22*32*5= 180. Now, I'll leave the hard part to you: how many distinct factors does 180 have?

HallsofIvy said:
In order that 10*N be a square, since 10= 2*5, N must have factors of 2 and 5. In order that 6*N be a cube, since 6= 2*3, N must have factors of 22 and 32. Since N is the smallest such number, it must be 22*32*5= 180. Now, I'll leave the hard part to you: how many distinct factors does 180 have?

Sorry, but the number of factors of 180 won't do him any good, since 10*180 is not a perfect square and 6*180 is not a perfect cube.

Halls, something you've let something slip there...

By the way Trailer Builder, the answer is ugly.

Werg22 said:
Halls, something you've let something slip there...

By the way Trailer Builder, the answer is ugly.