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28 Mar 2015, 22:39
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (01:49) correct
60%
(01:53)
wrong
based on 35
sessions
History
Date
Time
Result
Not Attempted Yet
How many trailing zeros will 11^50 - 1 will have ?
A) 3
B) 4
C) 5
D) 6
E) 7
A) 3
B) 4
C) 5
D) 6
E) 7
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29 Mar 2015, 04:25
Kudos
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CrazyIvan
i doubt this is a GMAT standard question , but let try to give it a shot .
\(11^5^0 - 1= (1+10)^50 -1 = 1+ ^5^0C_1 *10 + ^5^0C_2 *10^2 + ^5^0C_3 *10^3 + .... +^5^0C_50 *10^5^0 -1\)
\(50*10 (1+ \frac{49}{2*1}*10 + \frac{49*48}{3*2*1} * 10^2 + ..... + \frac{49*48...*2}{50!} 10^4^9)\)
taking LCM 50!
\(50*10 *( 50!+ \frac{49}{2*1}*10 *\frac{50!}{2!} + \frac{49*48}{3*2*1} * 10^2 *\frac{50!}{3!} + .....)/50!\)
we can take out 50 from the inner brackets and the remaining quantity will still be > 1
so answer should be 3 .
but i am not very sure .
29 Mar 2015, 10:28
Kudos
Bookmarks
It took me a lot of time to solve it, not only 2 minutes... so I wouldn't have been able to solve it in a test
I noticed that the powers of 11 can be calculated as
\(11^0=1 \\11^1=11 \\ 11^2=121 \\11^3=1331 \\11^4=14641\)
It reminds me of the Pascal's triangle, and the powers of 11 can be written as \(11^n=\sum_{i=0}^{n} (10^{i} \cdot \frac{n!}{i!(n-i)!})\)
I consider the last 4 digits, so I calculate \(1000\cdot \frac{50!}{47!3!}+100\cdot \frac{50!}{48!2!}+10 \cdot \frac{50!}{49!1!} +1=39200000+1225+500+1=39223001\)
If we subtract 1, there are three zeroes, so A is the correct answer.
Unfortunately I cannot explain it in a simpler way
I noticed that the powers of 11 can be calculated as
\(11^0=1 \\11^1=11 \\ 11^2=121 \\11^3=1331 \\11^4=14641\)
It reminds me of the Pascal's triangle, and the powers of 11 can be written as \(11^n=\sum_{i=0}^{n} (10^{i} \cdot \frac{n!}{i!(n-i)!})\)
I consider the last 4 digits, so I calculate \(1000\cdot \frac{50!}{47!3!}+100\cdot \frac{50!}{48!2!}+10 \cdot \frac{50!}{49!1!} +1=39200000+1225+500+1=39223001\)
If we subtract 1, there are three zeroes, so A is the correct answer.
Unfortunately I cannot explain it in a simpler way
