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# If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111)

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Re: If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111) [#permalink]
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f(x) = 90x^2 +20x +1

Adding and subtracting 10x^2 to above -
(100x^2 +20x +1) - 10x^2
(10x+1)^2 - 10x^2

Using a^2 - b^2 we get -
1*(20x + 1) and putting value of x as 11111
This gives final value as 222221. Sum of digits is 11 (Choice B)
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Re: If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111) [#permalink]
Why so complicated
Simple 90*X^2 = 9*10*some digit, hence sum of digits will definitely be 9(property of 9)
20X = 2*10*11111 = 222220 sum of digits 1
And 1
So 9+1+1 = 11
Hence B

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Re: If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111) [#permalink]
Shrutyforu wrote:
Why so complicated
Simple 90*X^2 = 9*10*some digit, hence sum of digits will definitely be 9(property of 9)
20X = 2*10*11111 = 222220 sum of digits 1
And 1
So 9+1+1 = 11
Hence B

Posted from my mobile device

­Shrutyforu which property of 9 are you talking about? Can you please explain a little bit?

Thank you!
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Re: If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111) [#permalink]
If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111)?

f(11111) = 90*(11111)ˆ2 + 20*11111 + 1 = 11111(999990 +20) + 1 = 11111*1000010 + 1 = 11111111111

Sum of the digits of f(11111) = 1+1+1+1+1+1+1+1+1+1+1 = 11*1 = 11

IMO B
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Re: If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111) [#permalink]
zlishz wrote:
Shrutyforu wrote:
Why so complicated
Simple 90*X^2 = 9*10*some digit, hence sum of digits will definitely be 9(property of 9)
20X = 2*10*11111 = 222220 sum of digits 1
And 1
So 9+1+1 = 11
Hence B

Posted from my mobile device

­Shrutyforu which property of 9 are you talking about? Can you please explain a little bit?

For Any multiple of 9, the sum of digits will ultimate add to 9...
Re: If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111) [#permalink]
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