Mind teaser Factorial Questions
Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100!
Q1. Find the reminder, when ‘a’ is divided by the product of first 7 natural numbers
a. 873
b. 746
c. 912
d. 413
e. None of the above.
Q2. Find the last two digits of ‘a’
a. 3
b. 9
c. 13
d. 19
e. 23
Q3. The number ‘a’ is ______ and ‘a/2’ is________.
a. Even, Even
b. Even, Odd
c. Odd, Odd
d. Odd, Even
e. None of the above
Q4. Find the remainder, when ‘a’ is divided by 168
a. 33
b. 129
c. 153
d. 67
e. 9
Q5. Number of digits in the number ‘a’ are
a. 137
b. 283
c. 314
d. 189
e. None of the above
Q1. Find the reminder, when ‘a’ is divided by the product of first 7 natural numbers
a. 873
b. 746
c. 912
d. 413
e. None of the above.
ans is a(873)
Little Star
n/a
Q4. Find the remainder, when ‘a’ is divided by 168
a. 33
b. 129
c. 153
d. 67
e. 9
ans is 33(a)
Little Star
n/a
Q3. The number ‘a’ is ______ and ‘a/2’ is________.
a. Even, Even
b. Even, Odd
c. Odd, Odd
d. Odd, Even
e. None of the above
last digit is 13 so a is odd ans when a divided by 2 so reminder 1 so a/2 also odd
Little Star
n/a
can u explain the solution in detail
n/a
just consider upto 9! as the 10! ends up with 2 0`s
=>13
n/a
hi can anyone explain solution in detail and what about ans of Q5 i think it should be none of these please do explain.
the number of numbers have a pattern
1+1+1+2+3+3+4+5+5+6+7+7+..................99+99+100
so use the abraic expression n(n-1)/2
then use An=n(n+1)/difference
Q4. 1!+2!+..........................+100!
sum of 1!+.....+6! = 873 as rest of the number will be divisible by 168(2*3*4*7) so divide 873 by 168 you will get 33 as remainder. If you have any doubt then just revert to me.
Q2. Find the last two digits of ‘a’
a. 3
b. 9
c. 13
d. 19
e. 23
ans is 13
Little Star
n/a
n/a