Question

This is your questions.
Today we practice only 3 type of questions which are asked in defence exams every year.

1st:-How many two-digits numbers are divisible by 4?
Answer. -The first digit which is divisible by 4 is 12 and last digit is 96
The common difference between the digits divisible by 4 is 4
so.A/q,
=96=12+(n-1)4
=96-12=(n-1)4
n=22
2nd -:if n! has 17 zeroes then what is the value of n?
sol-:For the occurance of zero in the result of any multiplication it should be that there is either a multiple of 10 or the multiplication of (2×5).
In the same way 20! will consist of 4 zeroes
Factorial-1st -:10!-2zeroes
2nd-:20!-4zeroes
3rd-: 30! -:6 zeroes
4th-:40!-:8zeroes
same like this 80!-:16 zeroes,(85!-:17 zeroes) answer.
What is the number of diagonals of an octagon?
Ans-:No. of diagonals in any polygon=n (n-3)÷2 - no. of sides=n
Octagon has=8 (8-3)÷2=20
Thanks

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Answer 1

Step-by-step explanation:

ok thanks for giving this

Answer 2

Answer:

thank you sir

It really helped me out..keep posting and help us please