Be observant! N cannot take 4 in St (2). All it can take is a positive integer ending in 2 or 8 only.theCodeToGMAT wrote:Kop, in my solution we have observed a pattern that starting from N=5 all the digits will have unit digit 0 because factors 2 & 5 will be present in all those factorials.
According to statement 2, N can be 2, 4, etc.. Because they will yield odd quotient .
Now, 1! +2!= 3
And, 1!+2!+3!+4!= 33
Both have 3 as unit digit.
Also, as we stated, 5! Onwards wil have unit digit as 0.. So, unit digit will be "3" only
Last Digit of factors
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sanju09
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Sanjeev K Saxena
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Sanjeev K Saxena
Quantitative Instructor
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www.manyagroup.com
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theCodeToGMAT
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Yeah correct!... instead of adding "1", mistakenly, i was doing subtract "1"sanju09 wrote: Be observant! N cannot take 4 in St (2). All it can take is a positive integer ending in 2 or 8 only.
..R A H U L
