let x is the number of the factors of 100

let y is the sum of the first 100 positive integers

let z is the number of zeroes in 100!

compute x+y+z

The answer is 5083.

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IT WILL BE BETTER IF YOU CAN CHANGE YOUR STATEMENT; SUM OF FIRST HUNDRED INTEGERS TO SUM OF FIRST 100 NATURAL NUMBERSSOLUTION:

(a) Number of factors of 100

=100 can be written as $2^{2}$

X$5^{2}$=Therefore, no. of factors

="+1" the power of prime numbers

=(1+2)(1+2)

= $\boxed{9}$

(b)1+2+3.....+99+100

= $\frac{n(n+1)}{2}$

= $\frac{100(101)}{2}$

= $\boxed{5050}$

(c)no. of trailing zeros:

= $\frac{100}{5}$ = $\boxed{20}$

= $\frac{20}{5}$ = $\boxed{4}$

(divide any given positive integer repeatedly by 5 until it gets divided by it to find the no. of trailing zeroes.)

therefore, the answer is:

= $\boxed{9}$ + $\boxed{5050}$ + $\boxed{20}$ + $\boxed{4}$

= $\boxed{5083}$