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14 May 2008, 02:16
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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:
35%
(medium)
Question Stats:
69% (01:28) correct
31%
(01:28)
wrong
based on 2052
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of \(1! + 2! + ... + 10!\)?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
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01 Feb 2011, 09:46
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Official Solution:
What is the value of \(1! + 2! + ... + 10!\)?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
For any integer \(n\) greater than or equal to 5, the units digit of \(n!\) is zero. This is because, for \(n \geq 5\), \(n!\) will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from \(5!\) through \(10!\) each have zero as their units digit. Adding up the remaining terms, we'd have \(1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33\). Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.
Alternatively, we can do the following:
\(1! + 2! + ... + 10! = 1 + (2! + ... + 10!)\) = odd + even = odd.
\(1! + 2! + ... + 10! = 3 + (3! + ... + 10!)\) = integer divisible by 3.
Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.
Answer: B
What is the value of \(1! + 2! + ... + 10!\)?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
For any integer \(n\) greater than or equal to 5, the units digit of \(n!\) is zero. This is because, for \(n \geq 5\), \(n!\) will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from \(5!\) through \(10!\) each have zero as their units digit. Adding up the remaining terms, we'd have \(1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33\). Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.
Alternatively, we can do the following:
\(1! + 2! + ... + 10! = 1 + (2! + ... + 10!)\) = odd + even = odd.
\(1! + 2! + ... + 10! = 3 + (3! + ... + 10!)\) = integer divisible by 3.
Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.
Answer: B
General Discussion
14 May 2008, 02:41
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sondenso
There's a easier way to crack this...
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
Now, all figures after 5! will have 0 in their unit digit (720 on 6!, 5040 on 7! etc.) so, all that we have to look into is the unit's place in the answer!
1+2+6+4 = 13 so, the answer should end with 3. Pick your answer!
