can u tell the numbers of zero at the right hand.?

Question

tell the number of zero at the end of the number

5^1 * 10 ^2 * 15^3 ---------100^20 / 5^22.

Answer 1

5^1 * 10 ^2 * 15^3 ---------100^20 / 5^22
= 5^(1+2+ ... ... + 20) x 1^1. 2^2 . 3^3 . 5^5 . .. . 10^10 . 15^15 . 20^20 / 5^22
= 5^(210 - 22) . 5^(5+15) . 10^10 . 10^20 .........
= 5^(210 -22+ 20) x 10^30 x .........
= 5^208 x 10^30 x ...
= 10^(208 +30) x ...
= 10^238 x ....

Ans: 238
I assumed sufficient number of 2's will be there to make your 5's >> 10
Please count yourself and write confirmed figure

Answer 2

Try simplify the equation by the following:

5^1 x 10^2 x 15^3 x 20^4 x 25^5 x 30^6 x 35^7 x 40^8 x 45^9 x 50^10 x 55^11 x 60^12 x 65^13 x 70^14 x 75^15 x 80^16 x 85^17 x 90^18 x 95^19 x 100^20 / 5^22

= 5^1 x 10^2 x (3^3 x 5^3) x (2^4 x 10^4) x (5^5 x 5^5) x (3^6 x 10^6) x (7^7 x 5^7) x (2^16 x 10^8) x (3^18 x 5^9) x (5^10 x 10^10) x (11^11 x 5^11) x (2^12 x 3^12 x 10^12) x (13^13 x 5^13) x (7^14 x 10^14) x (3^15 x 5^15 x 5^15) x (2^16 x 2^16 x 2^16 x 10^16) x (17^17 x 5^17) x (3^36 x 10^18) x (19^19 x 5^19) x (10^20 x 10^20) / 5^22

First count the factor of 10’s

2+4+6+8+10+12+14+16+18+20+20 = 130

Count the factor of 2’s & 5’s as the product of 2x5 = 10

2’s : 4+8+8+12+16+16+16 = 80

5’s : 1+3+5+5+7+9+10+11+13+15+15+17+19 = 130
the fact that denominator is 5^22 , so 130 – 22 = 108 which is still more than no. of 2’s, we can take 2^80 x 5^80 as no of factor of 10’s

Thus we can add 80 to 130 = 210,

The no’s of 0’s at the end of the number is 210.

Answer 3

Each trailing zero results from multiplying by 10. Multiplying by 10 means that you need a factor of 2 and a factor of 5. So to find the number of zeros, count how many factors of 2 and factors of 5 there are and which ever number is less, that will be the number of zeros.

Every number counting by 5s (5, 10, 15, 20...) has a factor of 5, that is the number is a multiple of 5. The numbers 25, 50, 75, and 100 each have two factors of five. When you exponentiate, you multiply the bnumber of factors by the exponent. So, for example 15^3 will have 3 factors of five, 25^5 will have 10. When you multiply numbers together the factors add so 5^1 * 10^2 * 15^3 has 1+2+3 = 6 factors of 5.

We know that 1+2+3...+(n-1) + n = (n+1)n/2 so (counting only single factors of 5 for now) 5^1 + 10^2 + 15^3 .... 100^20 has 1 + 2 + 3 + ...20 = 21*20/2 = 210 factors of five. Adding in the extra factors for 25^5, 50^10, 75^15, 100^20 gives 5 + 10 + 15 + 20 = 65 more factors of 5 for a total of 275. All we know so far is that 5^1 * 10^2 * 15^3 ... 100^20 has 275 factors of 5.

Next you divide by 5^22 which divides out 22 of the factors leaving 253 factors of 5.

Now you need to see how many factors of 2 there are. The odd fives (5, 15, 25, etc.) have none. The even fives (10, 20, 30, etc.) have at least one, the double evens (20, 40, 60, 80, 100) have at least 2 and 40 has 3 factors of 2 and 80 has 4. Adding these factors of 2 for 10^2 + 20^4 + 30^6 + 40^8 .... 100^20 is a matter of multiplying the number of factors of two in the base number by the exponent. This gives: 1*2 + 2*4 + 1*6 + 3*8 + 1*10 + 2*12 + 1*14 + 4*16 + 1*18 + 2*20 = 180

So your number has 252 factors of 5 but only 180 factors of 2. Since it takes one of each factor to produce a trailing zero, you will only have 180 trailing zeros.

Answer 4

10 to 100 shall count number of power of 2

so we have power of 2 = 2+4+6+... + 20+2 = 112
as 100 has 2^2 contributing
power of 5

upto 100 1+2+3... 20 = 20*21/2 = 205

now 25 50 75 100 give one extra 5 each

power of 5 = 209

devide by 5^22
so power of 5 = 187

so number of zeros at the end = 110(as number of 2 s = 110)
number of 5 = 187

so lower of the 2 or 112

Answer 5

It's 210 zeroes.

Answer 6

Looks like it's 202; you might want to double-check my counting though.