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Source: — Problem Solving |

by f2001290 » Mon May 28, 2007 12:09 am
We can say that r(s+1) is odd

Odd * Odd = Odd
So, r and (s+1) are odd

For (s+1) to be odd, s should be even
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Re: TestPrep

by jayhawk2001 » Mon May 28, 2007 7:59 am
moneyman wrote:Pls explain..

1)10^8-10^2/10^7-10^3=


What are the answer choices ?

10^2*[ 10^6-1] / 10^3*[10^4-1]

1 is small compared to 10^6 and 10^4.

10^2*10^6 / 10^3*10^4
= 10
moneyman wrote:
2)If r ans s are integers and rs+r is odd,which of the following must be even?
a)r
b)s
c)r+s
d)rs-r
e)r^2+s


Thanks..
r(s+1) is odd. So, s has to be even
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The answer

by moneyman » Mon May 28, 2007 8:25 am
The answer to the 1st question is 10..
Maxx
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by arunmmw » Sat Jul 12, 2008 11:16 pm
I am quite confused with the second question,

r(s+1) doesnt mean that s has to be positive, s can also be 0. My question is,
whether 0 is a positive integer?
Regards
Arun
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Re: TestPrep

by ildude02 » Sun Jul 13, 2008 8:40 am
Hi jayhawk,

Is it Ok to leave out the subtraction of 1 without any problem. Are there any considerations we need to take into account before doing so?

If an example says, 2 * 10 ^16 /10 ^4, and the answer asks for approximate answer; can we say it's 10 ^ 12 ? Especially if the ansewer choice even has 10 ^ 11 and 10 ^ 13 along with 10 ^ 12 choice.

Appreciaet your response.
jayhawk2001 wrote:
moneyman wrote:Pls explain..

1)10^8-10^2/10^7-10^3=


What are the answer choices ?

10^2*[ 10^6-1] / 10^3*[10^4-1]

1 is small compared to 10^6 and 10^4.

10^2*10^6 / 10^3*10^4
= 10
moneyman wrote:
2)If r ans s are integers and rs+r is odd,which of the following must be even?
a)r
b)s
c)r+s
d)rs-r
e)r^2+s


Thanks..
r(s+1) is odd. So, s has to be even
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by Ian Stewart » Sun Jul 13, 2008 4:11 pm
arunmmw wrote:I am quite confused with the second question,

r(s+1) doesnt mean that s has to be positive, s can also be 0. My question is,
whether 0 is a positive integer?
Zero is not a positive integer, and is not a negative integer. Zero is an even integer. Still, I'm confused by the question- we aren't asked here about positive or negative integers, we're asked about even and odd integers. When dealing with evens and odds, the same rules apply regardless of whether integers are negative or positive. If we know rs+r is odd, then we know r(s+1) is odd. The only way a product of integers can be odd is if all the terms in the product are odd: r is odd, and s+1 is odd. So r must be odd and s must be even. B is the correct answer.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: TestPrep

by Ian Stewart » Sun Jul 13, 2008 4:19 pm
ildude02 wrote: Is it Ok to leave out the subtraction of 1 without any problem. Are there any considerations we need to take into account before doing so?

If an example says, 2 * 10 ^16 /10 ^4, and the answer asks for approximate answer; can we say it's 10 ^ 12 ? Especially if the ansewer choice even has 10 ^ 11 and 10 ^ 13 along with 10 ^ 12 choice.
When we add very small things to very large things, we can normally ignore the small things when making an estimate. For example, 10^6 + 10^2 is very close to 10^6: we're adding one million to one hundred here, and one hundred is tiny compared to a million. In your example above, you are multiplying by two: you're doubling the result. 10^12 is one trillion, while 2*10^12 is two trillion. That's still closer to 10^12 than it is to 10^13 (ten trillion) or to 10^11 (100 billion), but it's not a very good estimate of 10^12 (one trillion)- it's twice as big.

That's a long way of saying that we often use this technique (of ignoring things that are minuscule relative to other terms) in estimating sums or subtractions, but we need to be more careful about products or quotients. Bear in mind as well that we're estimating the answer to the question in the Original Post; while the answer is very close to 10, it is not exactly equal to 10.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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