\(S=k*10^9= 10^9+2*11*10^8+3*11^2*10^7+........+10*11^9\)........(1)

\((\frac{11}{10})S= 11*10^8+2*11^2*10^7+........+9*11^9+11^{10}\)......(2)

Subtract (1) from (2)

\(\frac{S}{10} = 11^{10}- [ 10^9+11*10^8+11^2*10^7........+11^9]\)

S/10 = 11^{10}- [10^9{(11/10)^{10} - 1}/{(11/10)-1}]

\(\frac{S}{10} = 11^{10} -[10^{10} ((\frac{11}{10})^{10} - 1)]\)

\(\frac{S}{10} = 11^{10} -[11^{10} - 10^{10}]\)

\(S= 10^{11}\)

\(k*10^9= 10^{11}\)

\(k= 10^2 = 100\)

10^9 * { 1+ 2*(1.1)^1 + 3*(1.1)^2 +... + 10*(1.1)^9 } = 10^9 * k

Estimate lower-bound value of k:
= 10^9 * { 1+ 2*(1)^1 + 3*(1)^2 +... + 10*(1)^9 }
= 10^9 * {1+ 2+ ...+10}
= 10^9 * 55 = 10^9 * k
k > 55 ---> eliminate choices (A) 12.1 and (B) 44.1

Estimate reasonable value of k:
= 10^9 * { 1+ 2*(1)^1 + 3*(1)^2 + ... +10*(1)^9} * 1.1^6
= 10^9 * 55 * ~1.6
= 10^9 * ~88 = 10^9 * k
k is close to 100

Thus, FINAL ANSWER IS (C)

I have calculated the whole log. It comes down to 10 to the power 11. So the answer should be C. But do they ask such questions in GMAT? If they do, there must be some way of doing it within 2 minutes which I cannot even begin to search for!

If \((10)^9+2(11)^1(10)^8+3(11)^2(10)^7+…+10(11)^9=k(10)^9\), then k is equal to

A. 121/10
B. 441/100
C. 100
D. 110
E. 120
Multiplying given equation's both sides by \(10^{-9}\)
\((10)^9*10^{-9} + 2(11)^1(10)^8*10^{-9} + 3(11)^2(10)^7*10^{-9} + 4(11)^3(10)^6*10^{-9} + 5(11)^4(10)^5*10^{-9} + 6(11)^5(10)^4*10^{-9}+\)
\(+ 7(11)^6(10)^3*10^{-9} + 8(11)^7(10)^2*10^{-9} + 9(11)^8(10)^1*10^{-9} + 10(11)^9*10^{-9} = k(10)^9*10^{-9}\)

\(1 + 2*11*10^{-1} + 3(11)^2*10^{-2} + 4(11)^3*10^{-3} + 5(11)^4*10^{-4} + 6(11)^5*10^{-5} + 7(11)^6*10^{-6}+\)
\(+ 8(11)^7*10^{-7} + 9(11)^8*10^{-8} + 10(11)^9*10^{-9} = k\)

\(1 + 2*1.1^1 + 3*1.1^2 + 4*1.1^3 + 51.1^4 + 6*1.1^5 + 7*1.1^6 + 8*1.1^7 + 9*1.1^8 + 10*1.1^9 = k\)

Solving for k gives ~ 100
(There must be a better way to solve it)

Answer C.
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If \((10)^9+2 (11)^{1}(10)^{8}+3 (11)^{2}(10)^{7}+…+10 (11)^{9}=k (10)^{9}\), then k is equal to

\(10^9*( 1 +2 (\frac{11}{10})+3 (\frac{11}{10})^{2}+…+10 (\frac{11}{10})^{9}=k (10)^{9}\)

\(k= 1 +2 (\frac{11}{10})+3 (\frac{11}{10})^{2}+…+10 (\frac{11}{10})^{9}\)
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--> multiply both sides by \(\frac{11}{10}\):
\(k(\frac{11}{10})= (\frac{11}{10}) + 2(\frac{11}{10})^{2} + 3(\frac{11}{10})^{3} +...+ 10(\frac{11}{10})^10\)

And subtract \(k\) from \(k(\frac{11}{10})\):

\(k(\frac{11}{10} -1)= -1 -( (\frac{11}{10}) + (\frac{11}{10})^{2} + (\frac{11}{10})^{3} +...+ (\frac{11}{10})^{9} ) + 10(\frac{11}{10})^{10}\)
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\((\frac{11}{10}) + (\frac{11}{10})^{2} + (\frac{11}{10})^{3} +...+ (\frac{11}{10})^{9}\) - geometric sequence

\(Sum = \frac{(\frac{11}{10})*(\frac{11}{10})^9 -1)}{(\frac{11}{10}-1)}= 11* (\frac{11}{10})^9 -1)\)


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\(k(\frac{11}{10} -1)= -1 - \frac{11^{10}}{10^{9}} +11 + \frac{11^{10}}{10^{9}}= 10\)

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

Answer (C)

Since 11 = 1.1 x 10, we can rewrite the left hand side of the equation as:

10^9 + 2(1.1 x 10)(10^8) + 3(1.1^2 x 10^2)(10^7) + … + 10(1.1^9 x 10^9) = k(10^9)

10^9 + 2(1.1)(10^9) + 3(1.1^2)(10^9) + … + 10(1.1^9 )(10^9) = k(10^9)

Dividing the equation by 10^9, we have:

1 + 2(1.1) + 3(1.1)^2 + … + 10(1.1^9) = k → Eq. 1

We need to find the value of k. Without a calculator, we can multiply the equation by 1.1 to obtain:

1.1 + 2(1.1)^2 + 3(1.1)^3 + … + 10(1.1^10) = 1.1k → Eq. 2

Subtracting Eq. 2 from Eq. 1 (by combining terms with the same powers of 1.1), we have:

1 + 1.1 + 1.1^2 + 1.1^3 + … + 1.1^9 - 10(1.1^10) = -0.1k → Eq. 3

Now we see that all the terms (except the last term) form a geometric sequence (notice that the first term 1 can be written as 1.1^0).

Recall that the sum of a finite geometric series is:

S_n = a_1 * (1 - r^n) / (1 - r)

where a_1 is the first term, n is the number of terms and r is the common ratio.

So the sum of the 10 positive terms (i.e., 1, 1.1, 1.1^2, …, 1.1^9) is:

S_10 = 1(1 - 1.1^10)/(1 - 1.1) = (1 - 1.1^10)/(-0.1)

Substituting this back into Eq. 3, we have:

(1 - 1.1^10)/(-0.1) - 10(1.1^10) = -0.1k

Multiply the above equation by -0.1, we have:

1 - 1.1^10 + 1.1^10 = 0.01k

1 = 0.01k

100 = k

Answer: C
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