Find the no. of consecutive zeros at the end of the following no.
100!*200!
How to calculate the last rightmost non zero digit in 30!
Hello Vidisha,
Please find the solution below.
30! = 2^26 x 3^14 x 5^7 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1
Now the for the first non zero digits,lets first remove the terms which are causing the trail of zeroes at the end...which is 5^7 x 2^7....so u r left with....
2^19 x 3^14 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1
the last digit of this pdt will be 8 x 9 x 1 x 1 x 9 x 7 x 9 x 3 x9 = 8..
so the first non zero digit from the right will be 8.
Find the no. of consecutive zeros at the end of the following no.
1!*2!*3!*4!*5*................................*50!
Power of 5 in 1!*2!*3!*4!*5*................................*50! = 262. Thus number of zeros at the end of this num = 262.
power of 5 in 100!*200!=73. Thus number of zeros at the end of this num = 73.
The number of zeroes at the end on N
N = 4 x 8 x 12 . . . .1000
a. 11 b. 6 c. 16 d. 62
woww
Hello, Vidisha.nagpal
Please find the solution below,
N = 4*8*12*....*1000 = (4*1)*(4*2)*(4*3)*.....*(4*250) = 4250*(1*2*3*...250) = 2500*250!
Now 250! = 2244*3123*562*740*1124*1320*1714*1913*2310*298*318*376*......
Hence the number of zeroes at the end of N is 62.
Nice instruction, thanks
How many zeroes are present at the end of 25!+26!+27!+28!+30!?
Hi Samyak
25! ( 1 + 26 + 26 × 27 + 26 × 27 × 28 + 26 × 27 × 28 × 29 × 30 )
25! ( 1 + 26 + 26 × 27 + 26 × 27 × 28 + 26 × 27 × 28 × 29 × 30 )
25! has 6 trailing zeros and, the term inside the bracket is divisible by 5
Hence, 6 +1 , 7 trailing zeros .
What power of 8 exactly divides 25! ?
Highest power of 2 in 25! = [25/2] + [25/2^2] + [25/2^3] +........
12 + 6 + 3 + 1 = 22
So , the highest power of 8 in 25! = [22/3] = 7
For how many positive integers does n! End with exactly 100 zeroes?
Good
ok
tuyet voi
good
