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23 Jun 2019, 22:08
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:40) correct
41%
(02:36)
wrong
based on 225
sessions
History
Date
Time
Result
Not Attempted Yet
If f(x) = 90x^2 + 20x + 1, then find the sum of the digits of f(11111)?
A. 10
B. 11
C. 101
D. 111
E. 112
A. 10
B. 11
C. 101
D. 111
E. 112
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
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23 Jun 2019, 23:18
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\(f(x)=90x^2+20x+1\)
\(f(x)=10x(9x+2)+1=10x(9x+1+1)+1\)
\(f(11111)=(10*11111)*((9*11111)+1+1)+1\)
\(f(11111)=(111110)*(99999+1+1)+1=(111110)*(100000+1)+1\)
=1 1 1 1 1 0 0 0 0 0 0
+0 0 0 0 0 1 1 1 1 1 0
+0 0 0 0 0 0 0 0 0 0 1
-----------------------------
=1 1 1 1 1 1 1 1 1 1 1
Sum of digits=11
Answer is (B)
\(f(x)=10x(9x+2)+1=10x(9x+1+1)+1\)
\(f(11111)=(10*11111)*((9*11111)+1+1)+1\)
\(f(11111)=(111110)*(99999+1+1)+1=(111110)*(100000+1)+1\)
=1 1 1 1 1 0 0 0 0 0 0
+0 0 0 0 0 1 1 1 1 1 0
+0 0 0 0 0 0 0 0 0 0 1
-----------------------------
=1 1 1 1 1 1 1 1 1 1 1
Sum of digits=11
Answer is (B)
General Discussion
19 May 2024, 13:39
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nick1816
The expression can be rearranged to:
(10X+1)(9X+1) + X
Substituting 11111:
(111110+1)(99999+1) + 11111
(111111)(100000) + 11111
11111100000
+
11111
_______________
11111111111 = 11 1's
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