Can someone assist on the bellow Question please?
A skeet shooting competition awards prizes as follows: the first place winner receives
11 points, the second place winner receives 7 points, the third place finisher receives 5
points, and the fourth place finisher receives 2 points. No other prizes are awarded.
John competes in the skeet shooting competition several times and receives points
every time he competes. If the product of all of the points he receives equals 84,700,
how many times does he participate in the competition?
GMAT Forum
4 postsPage 1 of 1
- Abdulla
- Eric_J
NP Ch1 Q:16.
Abdulla -
I've never posted a response on here before, but I'll give it a try.
I think the first thing you are supposed to notice is that all the possible scores are prime numbers.
11, 7, 5 and 2 are all prime and cannot be expressed as the product of smaller integers. The question then gives you the product of John's scores - I think the context of the question here is just to throw you for a loop - normally we think of the total or "sum" when we think of scores, but they've given us the product here (i.e., all the scores multiplied together). we know that product is 84,700. if we can factor 84,700 back into the individual scores that it is composed of, we can know how many times John received each score.
so: 84700
847 X 100
847 X 10 X 10
847 X 2 X 5 X 2 X 5 (breaking each '10' into 2X5)
7 X 121 X 2 X 5 X 2 X 5 (breaking 847 into 7 X 121)
7 X 11 X 11 X 2 X 5 X 2 X 5 (breaking 121 into 11 X 11)
all of those numbers are prime, so re-ordering them
11 X 11 X 7 X 5 X 5 X 2 X 2 = 84,700
so we know John got two 11's, one 7, two 5's and two 2's. we also know from the question that every time he competed he won a prize. so, if he won 7 prizes (two + one +two + two), then he competed 7 times.
Does that help?
Eric
I've never posted a response on here before, but I'll give it a try.
I think the first thing you are supposed to notice is that all the possible scores are prime numbers.
11, 7, 5 and 2 are all prime and cannot be expressed as the product of smaller integers. The question then gives you the product of John's scores - I think the context of the question here is just to throw you for a loop - normally we think of the total or "sum" when we think of scores, but they've given us the product here (i.e., all the scores multiplied together). we know that product is 84,700. if we can factor 84,700 back into the individual scores that it is composed of, we can know how many times John received each score.
so: 84700
847 X 100
847 X 10 X 10
847 X 2 X 5 X 2 X 5 (breaking each '10' into 2X5)
7 X 121 X 2 X 5 X 2 X 5 (breaking 847 into 7 X 121)
7 X 11 X 11 X 2 X 5 X 2 X 5 (breaking 121 into 11 X 11)
all of those numbers are prime, so re-ordering them
11 X 11 X 7 X 5 X 5 X 2 X 2 = 84,700
so we know John got two 11's, one 7, two 5's and two 2's. we also know from the question that every time he competed he won a prize. so, if he won 7 prizes (two + one +two + two), then he competed 7 times.
Does that help?
Eric
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
eric_j for the win.
that's a spectacular explanation, to which i can add little of value. instead, i'll just add the following general takeaway:
GENERAL TAKEAWAY: if a probelm specifies the product of some numbers, then you should IMMEDIATELY find the prime factorization of the number that's given as product.
that prime factorization will provide the heart of the information you need to solve the problem. you may have to find some crazy new way of manipulating that information, but you will need the prime factorization.
that's a spectacular explanation, to which i can add little of value. instead, i'll just add the following general takeaway:
GENERAL TAKEAWAY: if a probelm specifies the product of some numbers, then you should IMMEDIATELY find the prime factorization of the number that's given as product.
that prime factorization will provide the heart of the information you need to solve the problem. you may have to find some crazy new way of manipulating that information, but you will need the prime factorization.
4 posts Page 1 of 1