Here is a good problem to test your understanding of prime factors, not to add multiples. See if you can finish the problem in less than 2 minutes.

If P is the product of all of the positive multiples of 11 less than 100, then what is the sum of the distinct prime factors of P?

(A) 22

(B) 28

(C) 45

(D) 49

(E) 89

Approach: The first way to crack this problem is to figure out what the multiples of 11 are. Start with the lowest and begin writing them on your scratch paper. Doing so will help you map out the problem.

11, 22, 33… by the time you get to 33, you might want to stop and think what the range of the multiples is. Meaning, what is the highest multiple of 11 that is still less than 100. The answer is 99 = 11 X 9. Therefore there are 9 multiples of 11, starting with 11 and ending with 99.

The crux of the problem—at least in terms of saving time—is how to factor out the primes of P. Note that 11 x 22 x 33…x 99 is going to give you a really large number. How do we make it easy to look for the prime factors?

One way is to see which of the primes (2, 3, 5, 7, 11, 13 etc.) we can extract from 11 x 22 x 33…x 99. A quicker way is to factor out the 11 so that the multiples of 11, up until 99, can then be written as P = 11^{9} (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9).

Note that the problem asks us for the distinct primes, so we only need to pull one of each prime from P. The first prime to notice is 11. There were nine of these in P but because we only need one of each distinct prime, we don’t have to worry about the exponent.

P = 11 (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9).

Now we just need to add up 11 + 2 + 3 + 5 + 7 and we get 28. Answer (B).

These types of problems are time-consuming. Yet, there are always ways to make a problem less laborious. In the problem above, we looked for a pattern once we started writing out the numbers. Remember, if you are writing out all of the terms of the sequence that will cost you too much time. On other hand if you do not write out anything at all then you will not move forward on the problem. At the end of the day, the long way is better than the no way (unless of course you are running out of time.)

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