 # Extra Hard SAT Math Question – Prime Factorization

At the end of every SAT Math section, the test makers try to come up with an extremely difficult problem that will leave even the cleverest students scratching their heads. The really evil part, though, is that even these problems can be solved in under a minute without a calculator – if you know what to do. This means that once you “figure out the trick,” these difficult problems become easy. So, while those test makers are busy cackling with sadistic glee, let’s see if we can’t beat them at their own game.

Consider the following brain teaser:

What is the least common multiple of 18 and 14 that is also a perfect square?

A simple enough question. But how do you figure it out? Could we just multiply 18 by 14?

18*14 = 252

That’s definitely a common multiple of 18 and 14, but is it a perfect square?

252^(1/2) = 15.87450787…

Should’ve known it wouldn’t be that easy. This IS a hard problem after all. Maybe we should multiply 18^2 by 14^2:

(18^2)*(14^2) = 63504

Well, that works, but is it the LEAST common multiple? Don’t look at me! How should I know? Hmm…something tells me it probably isn’t. Maybe we could just go through the multiples of 18 trial and error style?

1*18 = 18. Not divisible by 14, not a perfect square.

2*18 = 36. It is a perfect square, but not divisible by 14.

3*18 = 54. Not divisible by 14, not a perfect square.

4*18 =

This is clearly going to take too long. We’re running out of time on this section – whatever shall we do? Never fear – prime factorization is here! Remember, all SAT math problems can be solved in under a minute without a calculator. If you know what to do, hard problems like this one become fast and easy. In this case, we should begin by finding the prime factors of both 18 and 14:

18 = 9*2 = 3*3*2

14 = 7*2

Remember, to find the prime factors of a number means to write out the numbers as products of only prime numbers like 2, 3, 5, 7, 11, 13, etc. In this case, the prime factors of 18 are 3, 3, and 2, and the prime factors of 14 are just 7 and 2. Now, if we just wanted to find the least common multiple of 18 and 14, we would multiply all the prime factors of 18 by all the prime factors of 14, but leaving out any duplicates that appear in both lists. In this case you would multiply:

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3*3*2*7 = 126

This works, because

(3*3*2)*7 = (18)*7 = 126

3*3*(2*7) = 3*3*(14) = 126

Note that you need only one 2 in 3*3*2*7, since both the prime factors of 18 (3*3*2) and 14 (7*2) include a 2. This is all well and good, except that 126 is not a perfect square:

126^(1/2) = 11.22497216…

So, what are we to do? Note that a perfect square multiplied by a perfect square is also a perfect square:

4*25 = 100

This means that we can pair up all of the unique prime factors of 126 so that we have a bunch of squares multiplied by each other. Thus,

3*3*2*7 = 126

becomes

(3*3)*(2*2)*(7*7) = 1764

We have to add an extra 2 and an extra 7 in order to make sure those numbers are part of perfect square pairs (we don’t need to add any extra 3s because there are already two of them). Sure enough, 1764 is a perfect square:

1764^(1/2) = 42

and 1764 is a multiple of both 18 and 14:

1764/18 = 98

1764/14 = 126

But, how can we be 100% sure that 1764 is the least common multiple? Well, think about it. Every perfect square that isn’t the square of a prime number will always be prime factorized into pairs like the ones above ((3*3)*(2*2)*(7*7) = 1764). Consider the following examples:

16 = 4*4 = (2*2)*(2*2)

36 = 6*6 = (3*2)*(3*2) = (3*3)*(2*2)

144 = 12*12 = (3*2*2)*(3*2*2) = (3*3)*(2*2)*(2*2)

Et cetera. It’s inevitable. If the prime factors of a perfect square aren’t two prime numbers, then they will consist of multiple pairs of other prime numbers. Now, is there any way we could remove any pairs or make any of these factors smaller without breaking the conditions of finding the least common multiple of 18 and 14 that is also a perfect square?

(3*3)*(2*2)*(7*7) = 1764

I don’t think so. Remove or change one number and the product will either no longer be a perfect square or no longer be a multiple of both 18 and 14. Thus, 1764 must be the least common multiple of both 18 and 14 that is also a perfect square.

To sum up, the “trick” to solving a problem like this is to:

-find the prime factors of the two numbers in question

-pair them up

-multiply

If you know what to do, it takes about 30 seconds to solve this problem. So you see, with practice, even the hardest problems on the SAT become easy. Check back here each week for more extra hard problems and the tricks you need to solve them! Also, remember that you can find out all the tricks from experts like me with a Test Masters course or private tutoring. Until  then, keep up the good work and happy studying!