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Lisapizzola
Can someone help me out with this?

What is the greatest integer k for which \(32.45 × 10^k\) is less than 1 ?

a. –2

b. –1

c. 0

d. 1

e. 2

Rule in exponents - \(10^x = \frac{1}{10^{-x}}\)

We need to find the greatest integer k such that \(32.45 × 10^k < 1\) -> \(\frac{32.45}{10^{-k}} < 1\)

This inequality becomes \(32.45 < 10^{-k}\) - Lowest value for inequality to hold good is k = -2

Therefore, the lower possible value for k such that \(32.45 × 10^k < 1\) is when k = -2(Option A)
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Solution



Approach and Working:
• \(32.45*10^k<1\)
• \(32.45<\frac{1}{10^k}\)
• \(32.45<10^{-k}\)
o Since 32.45 is less than 100, \(10^{-k}= 10^2\)
o k=-2


Hence, the correct answer is option A.

Answer: A
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Hello from the GMAT Club BumpBot!

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