Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225
C
Find the number of trailing zeros in the product of
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(1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).
if you observe the number are multiples of 5 and are power of the same number.
therefore the expression is as
(1^1)*(5^5)*(10^10)*(15^15)*(20^20)*(25^25)*(35^35)*(40^40)*(45^45)*(50^50).
=> (1^1)*(5^5)*(10^10)*(3^15*5^15)*(2^20*10^20)............(5^50*10^50)
now all the ten terms that can be formed are
10^10*10^20*10^30*10^40*10^50*(5^100*2^100)
=>10^250
if you observe the number are multiples of 5 and are power of the same number.
therefore the expression is as
(1^1)*(5^5)*(10^10)*(15^15)*(20^20)*(25^25)*(35^35)*(40^40)*(45^45)*(50^50).
=> (1^1)*(5^5)*(10^10)*(3^15*5^15)*(2^20*10^20)............(5^50*10^50)
now all the ten terms that can be formed are
10^10*10^20*10^30*10^40*10^50*(5^100*2^100)
=>10^250
nasheen wrote:Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225
C
We have a trailing zero when we 2 is multiplied by 5. In the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there are more 5-s than 2-s so the number of 2's will be deciding factor for the number of trailing zeros.
If we factor 2's then it is 10 ^ 10, 20 ^ 20, 30 ^ 30, 40 ^ 40, 50 ^ 50
= 2 ^ (10+40+30+120+50) * (Remaining) = 2 ^ (250) * (Remaining)
So there will be 250 trailing zero
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I dint got these concept of trailing zeroFLUID wrote:nasheen wrote:Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225
C
We have a trailing zero when we 2 is multiplied by 5. In the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there are more 5-s than 2-s so the number of 2's will be deciding factor for the number of trailing zeros.
If we factor 2's then it is 10 ^ 10, 20 ^ 20, 30 ^ 30, 40 ^ 40, 50 ^ 50
= 2 ^ (10+40+30+120+50) * (Remaining) = 2 ^ (250) * (Remaining)
So there will be 250 trailing zero
Why we need to multiply 2 by 5?