HW Bank 2.2.10

[jeremy.hansen]

I am a bit confused and I am hoping someone can put this into plain language for me

Q:If g has both 21 and 30 as factors, then what is the largest number that g must be a multiple of?

Where I go wrong is in using prime factors multiple times

21= 7*3
35=5*7

7*3*5*7 = 735

But the answer states

21 = 3 × 7. So f must have 3 and 7 as prime factors
30 = 2 × 3 × 5. So f must have 2, 3 and 5 as prime factors.

Combined: f must have 2, 3, 5 and 7 as prime factors. (Notice, you don't "double-count" the 3's.)

Why do you not double count? What in the question should trigger me to know not to double count?

Could the question be worded so that using both 7's would be appropriate?

Thanks

[Ben Ku]

Why do you not double count? What in the question should trigger me to know not to double count?

Could the question be worded so that using both 7's would be appropriate?

Thanks

The issue of double counting is not in the "wording" but rather in the concepts.

First, let me try to illustrate that your understanding is not correct. Another way to write the question is "What is the largest number g must be divisible by?" You're suggesting that the largest number would be 21*30 or 3*7*2*3*5 = 630. So does every number that have 21 and 30 as factors divisible by 630?

Well 210, 420, 630, 840 are all have 21 and 30 as factors, but only 630 is divisible by 630. However, all of these numbers are divisible by 210, or 2*3*5*7.

I think the way to think about it is, if 21 is a factor of g, that means that g must have 3 and 7 as factors. If 30 is a factor of g, then g must have 2, 3, and 5 as factors. However, the "3" from 21 and the "3" from 30 could possibly be the same. So to avoid "double counting" we would just count it once.

If the problem instead stated that g had 63 and 30 as factors, then, because 63 = 3*3*7, we know g must have two "3"s as factors, and we would count both of them. However, we would not count the "3" from 30 as additional.

I hope that makes sense.