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Re: Is 5^k less than 1,000? [#permalink]
23 Feb 2012, 23:54

1

This post received KUDOS

Expert's post

Is 5^k less than 1,000?

Is 5^k<1,000?

(1) 5^(k+1) > 3,000 --> 5^k>600 --> if k=4 then the answer is YES: since 600<(5^4=625)<1,000 but if k=10, for example, then the answer is NO. Not sufficient.

(2) 5^(k-1) = (5^k) - 500 --> we can solve for k and get the single numerical value of it, hence this statement is sufficient. Just to illustrate: 5^k-5^{k-1}=500 --> factor out 5^{k-1}: 5^{k-1}(5-1)=500 --> 5^{k-1}=125 --> k-1=3 --> k=4. Sufficient.

Re: Is 5^k less than 1,000? [#permalink]
11 Dec 2013, 21:24

Bunuel wrote:

Is 5^k less than 1,000?

Is 5^k<1,000?

(1) 5^(k+1) > 3,000 --> 5^k>600 --> if k=4 then the answer is YES: since 600<(5^4=625)<1,000 but if k=10, for example, then the answer is NO. Not sufficient.

(2) 5^(k-1) = (5^k) - 500 --> we can solve for k and get the single numerical value of it, hence this statement is sufficient. Just to illustrate: 5^k-5^{k-1}=500 --> factor out 5^{k-1}: 5^{k-1}(5-1)=500 --> 5^{k-1}=125 --> k-1=3 --> k=4. Sufficient.

Answer: B.

Hope it's clear.

Hey Bunuel,

could you explain why you can factor out 5^k-1 from 5^k? I don't understand why that is possible.

Re: Is 5^k less than 1,000? [#permalink]
12 Dec 2013, 02:20

Expert's post

unceldolan wrote:

Bunuel wrote:

Is 5^k less than 1,000?

Is 5^k<1,000?

(1) 5^(k+1) > 3,000 --> 5^k>600 --> if k=4 then the answer is YES: since 600<(5^4=625)<1,000 but if k=10, for example, then the answer is NO. Not sufficient.

(2) 5^(k-1) = (5^k) - 500 --> we can solve for k and get the single numerical value of it, hence this statement is sufficient. Just to illustrate: 5^k-5^{k-1}=500 --> factor out 5^{k-1}: 5^{k-1}(5-1)=500 --> 5^{k-1}=125 --> k-1=3 --> k=4. Sufficient.

Answer: B.

Hope it's clear.

Hey Bunuel,

could you explain why you can factor out 5^k-1 from 5^k? I don't understand why that is possible.

Operations involving the same bases: Keep the base, add or subtract the exponent (add for multiplication, subtract for division) a^n*a^m=a^{n+m}