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If the product of the integers a, b, c, and d is 546...

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by diebeatsthegmat » Sun Nov 07, 2010 11:25 am
phoenixhazard wrote:If the product of the integers a, b, c, and d is 546 and if 1<a<b<c<d, what is the value of b+c?
a) 273
b) 185
c) 21
d) 10
e) 4

OA = D
ohh you know abcd=546
546=2*3*7*13
so 3+7=10
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by phoenixhazard » Sun Nov 07, 2010 11:26 am
diebeatsthegmat wrote:
phoenixhazard wrote:If the product of the integers a, b, c, and d is 546 and if 1<a<b<c<d, what is the value of b+c?
a) 273
b) 185
c) 21
d) 10
e) 4

OA = D
ohh you know abcd=546
546=2*3*7*13
so 3+7=10
I understand that but can you explain how you got 2,3,7, and 13? I can't just think of those 4 numbers off the top of my head to multiply to 546, did you use a trick?
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by ikaplan » Sun Nov 07, 2010 12:39 pm
Whenever you see such a problem, think of divisibility issues.

First, rephrase the question: The problem states that 546 is composed of at least 4 factors and also points out to the relationship of those factors (a<b<c<d)

Therefore, start testing the divisibility of 546 with the smallest possible factor (in this case 2). 546/2=273
Then, check if 273 is divisible with 3: 273/3=91 (in case 273 was not divisible with 3, check if it is divisible with 4).

Then, "c" is less than "d" which indicates that you should split 91 among two factors. Check if 91 is divisible with 4, 5, 6 etc.

You will arrive to the conclusion that 91 is divisible with 7 (91/7=13) so c is 7 and d is 13.

This problem involves some "plug-in and play" exercise but if you know the divisibility rules, it is easy.

In conclusion, b+c=3+7=10

My answer is (D).
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