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grockit factors

by bblast » Sat May 28, 2011 11:25 am
How many different positive even integers are factors of 196?

1

2

3

5

6


oa-6

[spoiler]as a shortcut can we solve by this method :
1>even factors - > 2^2*7---no of even factors = 3*1=3
2>even factors - > 2*7^2---no of even factors = 1*3=3
total=6[/spoiler]
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by maihuna » Sat May 28, 2011 11:34 am
196 = 4*49 = 2^2*7^7
diff evens will be : 2 for 2^2
other factors are : 2 each for 7 and 7^2 i.e 4

a total of 6.

easy way to see : 2^2*7^2 have a total of 9 factors only 1, 7, 49 will be not even. :(
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by edvhou812 » Sat May 28, 2011 5:11 pm
196 = 4*49
196=98*2

Even integers seen: 2, 4, and 98. So there are three positive even integers that are factors of 196. Am a bit confused as to why the answer is six. However I am also puzzled as to whether or not to count reoccurring integers multiple times for these types of problems.
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by GMATGuruNY » Sat May 28, 2011 5:33 pm
edvhou812 wrote:196 = 4*49
196=98*2

Even integers seen: 2, 4, and 98. So there are three positive even integers that are factors of 196. Am a bit confused as to why the answer is six. However I am also puzzled as to whether or not to count reoccurring integers multiple times for these types of problems.
You're overlooking some even factors.

196 = 2*2*7*7.
We must count every combination of these factors that yields an even product:
2
2*2=4
2*7=14
2*2*7=28
2*7*7=98
2*2*7*7=196.
Thus, there are 6 even factors: 2,4,14,28,98,196.

I would solve the same way as bblast.

To determine the number of factors of an integer:

1. Prime-factorize the integer
2. Add 1 to each exponent
3. Multiply

196= 2^2 * 7^2.
Adding 1 to each exponent and multiplying, we get (2+1)(2+1)=9 total factors.
Odd factors are 1,7,49 = 3 factors.
Thus, the total number of even factors = 9-3=6.

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by bblast » Sat May 28, 2011 11:42 pm
Hi Mitch,
A small doubt. M taking a shortcut, is this correct ?

196 = 2 square * 7 square.

number of even factors = (number of ways of selecting 1 or 2 two's)*(number of ways of selecting 0 or 1 or 2 seven's)

i.e 2*3 = 6 even factors. I guess this is right.

If we had to find odd factors :

odd = (number of ways of selecting 0 out of 2 two's)*(number of ways of selecting 0 or 1 or 2 seven's)

i.e 1*3 = 3 odd factors.
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