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Math Expert V
Joined: 02 Sep 2009
Posts: 65184
If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 29% (02:02) correct 71% (02:27) wrong based on 45 sessions

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Competition Mode Question

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

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Director  V
Joined: 30 Sep 2017
Posts: 965
GMAT 1: 720 Q49 V40 GPA: 3.8
If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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4
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

Originally posted by chondro48 on 12 Mar 2020, 01:50.
Last edited by chondro48 on 13 Mar 2020, 17:21, edited 4 times in total.
Director  D
Joined: 25 Jul 2018
Posts: 731
Re: If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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

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

$$k= 1 +2 (\frac{11}{10})+3 (\frac{11}{10})^{2}+…+10 (\frac{11}{10})^{9}$$
---------------------------------------------------------------------------------------------------------------------------------------------
--> 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 $$k(\frac{11}{10})$$:

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

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

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

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

DS Forum Moderator V
Joined: 19 Oct 2018
Posts: 1996
Location: India
If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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1
1
$$S=k*10^9= 10^9+2*11*10^8+3*11^2*10^7+........+10*11^9$$........(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}]

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

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

$$S= 10^{11}$$

$$k*10^9= 10^{11}$$

$$k= 10^2 = 100$$
Sloan MIT School Moderator V
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Re: If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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1
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)

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Joined: 27 May 2018
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GMAT 1: 730 Q48 V42 Re: If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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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!
Target Test Prep Representative V
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Re: If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  [#permalink]

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Bunuel wrote:

Competition Mode Question

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

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

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# If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + … + 10(11)^9 = k(10)^9  