KCET · Maths · Binomial Theorem
The
\(1 !+4 !+7 !+10 !+12 !+13 !+15 !+16 !+17 ! \quad\) is divisible by
- A 4
- B 3 !
- C 5
- D 7
Answer & Solution
Correct Answer
(B) 3 !
Step-by-step Solution
Detailed explanation
As we know the last two digits of 10 ! and above factorials will be zero-zero.
\(\therefore 1 !+4 !+7 !+10 !+12 !+13 !+15 !+16 !+17 !\)
\(=1+24+5040+10 !+12 !+13 !\)
\(+15 !+16 !+17 !\)
\(=5065+10 !+12 !+13 !+15 !+16 !+17 !\)
In this series, the digit in the ten place is 6 which is divisible by \(3 !\).
\(\therefore 1 !+4 !+7 !+10 !+12 !+13 !+15 !+16 !+17 !\)
\(=1+24+5040+10 !+12 !+13 !\)
\(+15 !+16 !+17 !\)
\(=5065+10 !+12 !+13 !+15 !+16 !+17 !\)
In this series, the digit in the ten place is 6 which is divisible by \(3 !\).
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