Pinaki Biswas
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There are 2 zeros in 10! as, 2*5 has one zero and 10*1 has another.
Using the same logic, we will get that 20! has 4 zeros.
However, 30! has 7 zeros because 5*5=25.
So, we have to look out for 5^x and multiples of them.
40! has 9 zeros.
50! has 11 + 1 zeros because 50=2*5^2.
60! has 14 zeros.
70! has 16 zeros.
80! has 18+1 zeros as 75=3*5^2.
90! has 21 zeros.
100! has 23+1 zeros =24 as 100=4*5^2.
We can find an algorithm:100/5=20
100/5^2=4
20+4=24; It is the number of zeros in 100!.
So now we have a powerful weapon to solve this problem.
No. of zeros in 1000!:1000/5=200
1000/5^2=40
1000/5^3=8
1000/5^4=1.6(We will consider the whole no. part only)
So no. of zeros in 1000!=249.
Similarly no. of zeros in 1100!=220+44+8+1=273.
1200!=240+48+9+1=298
1150!=230+46+9+1=286
1175!=235+47+9+1=292
1170!=234+46+9+1=290.
We got our answer. It is 1170. 🙂
Proof:1. If there is x in a set, then it will have to be in a partition.(symmetric)
2.If x ~y, they are in the same partition.
=>y~x(reflexive)
3.If x~y,y~z,
Then x~z as all are in the same partition.(transitive)
Sir, here is the proof:
Given: The number is an INTEGER.
To Prove: If a number is divisible by 3, then the SUM of the DIGITS is divisible by 3.(Converse)
Proof: Suppose there exists a number abc with a as HUNDREDS PLACE, b as TENS and c as ONES.
- abc= 100a+10b+c
Suppose a+b+c=3n where n is an INTEGER.
Divide 100a+10b+c by 3.
You will get the REMAINDER as a+b+c.
a+b+c is divisible by 3.
Therefore, 100a+10b+c is DIVISIBLE by 3.
So, abc is DIVISIBLE by 3. (Proved) 🙂
Given : AB+AD+AC=AB+BC+BD=BC+CD+AC=BC+AD+BD
To Prove : Quadrilateral ABCD is a rectangle.
Construction : None required.
Proof : We know that AB+AD+AC=AB+BC+BD=BC+CD+AC=CD+AD+BD.
Thus, AB+AD=BC+CD; (1)
AB+BC=CD+AD. (2)
From subracting (2) from (1), we get:
AB-CD=CD-AB.
So,AB=CD.
Similarly,BC=AD.
Thus, ABCD is a parallelogram... (The proof is not completed yet)
Since: AB=BC,
BC is common,
AC=BD,
Angles ABC=BCD.
By proving all the angles equal by the method of congruency above,
All angles must be right angles.
Thus, ABCD is a rectangle. (Proved)
Given : AB+AD+AC=AB+BC+BD=BC+CD+AC=BC+AD+BD
To Prove : Quadrilateral ABCD is a rectangle.
Construction : None required.
Proof : We know that AB+AD+AC=AB+BC+BD=BC+CD+AC=CD+AD+BD.
Thus, AB+AD=BC+CD; (1)
AB+BC=CD+AD. (2)
From subtracting (2) from (1), we get:
AB-CD=CD-AB.
So,AB=CD.
Similarly,BC=AD.
So, BC=AD.
Thus, ABCD is a parallelogram... (The proof is not completed yet)
Since: AB=BC,
BC is common,
AC=BD,
Angles ABC=BCD.
By proving all the angles equal by the method of congruency above,
All angles must be right angles.
Thus, ABCD is a rectangle. (Proved)