If the units digit of 5,610.37/10^k is 6, what is the value of k?

Useful property: \(x^{-n}=\frac{1}{x^n}\)

We can also write the property in reverse order: \(\frac{1}{x^n}= x^{-n}\)

This means we can take: \(\frac{5,610.37}{10^k}\) and rewrite it as: \(5,610.37 \times 10^{-k}\)

Since we want the unit's digit to be 6, we want our resulting decimal to be 56.1037

In other words: \(5,610.37 \times 10^{-k} = 56.1037\)

So our goal is to move the decimal place two spaces to the left.

We can accomplish this by multiplying \(5,610.37\) by \(10^{-2}\)

In other words: \(5,610.37 \times 10^{-2} = 56.1037\)

From this we can conclude: \(-k = -2\), which means \(k = 2\)

Answer: B

Cheers,
Brent
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Not sure but I guess the units tens hundred digit is countable before decimals only.
So answer is supposedly 2=k
5610.37 {here unit digit is 0}
In order to make 6 as unit digit we need 100 in denominator
So, 5610.37/100=56.1037
100=10^k
10^2=10^k
K=2
IOC is B

Posted from my mobile device
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Since the Denominator is 10 raised to power k
From identity of exponents number raised to power of 10 in denominator , the decimal points shifted towards left of the numerator
Similarly, if multiplied by 10 raised to some number, the decimal point shifted towards right.
So answer is (B)=2

Without doing any math, since we are told of unit digit which we will get if the decimal point was moved 2 spaces to the left. Thus the answer should be B - 2 .

we can re write the expression \(\frac{5,610.37}{10^k}\)
\(\frac{5,61037}{10^ (k+2)}\)
since unit digit we get is 6 ; so k has to be 2
OPTION B ; 2

If we move the decimal point of 5610.37 two places to the left, then the resulting number will be 56.1037, which has a units digit of 6.

We see that if k = 2, then 5610.37 / 10^2 = 56.1037 has a units digit of 6. So k must be 2.

Answer: B
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This is a simple question.

If the decimal has to be right next to 6. Then the decimal has to shift left. If the decimal has to shift left then the value of k has to be negative.

In this option D and option E are eliminated.

If you observe the decimal has to shift left twice, in that case the value of k has to be negative 2.


OR

If we have to solve mathematically

5610.37 x \(10^{-k}\) = 56.1037

\(\frac{561037}{100}\) x \(10^{-k}\) = \(\frac{561037}{10000}\)

\(10^{-k}\) = \(\frac{1}{100}\)

\(10^{-k}\) = \(10^{-2}\)

therefore -k = -2 => k = 2
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Answer: Option B

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