Find the number of zeroes in product 1^1 * 2^2 * 3^3 * 4^4*……………
Soln.: Summetion of all powers of 5 denotes the no. of zeros at d end... 5^5*10^10*15^15*20^20*25^25...*100^100 5+10+15+20+50(for 25^25)+30+...+200 By visual inspection Total powers of 5: (5+10+...+100)(for powers of 5)+(25+50+75+100)(for powers of 25) =5*20*21/2+25*4*5/2 =1050+250 =1300
So option (b) is correct....
Give your best to the world. And the best will come back to you....
Hi can you please give some more insight on this explanation,please explain it in detail. this funda is not clear to me i have seen some questions on this site like how many zeroes will be in product 15*30*45*.....................1500 it is given there 15^20*!50 i understand !50 but how come 15^20 please explain all this and clear my doubt. thanks in advance
Hi guys I am trying to keep this thing understanble..Pls go thru this one..
As we want to find the no. of zeros(in base 10) so we can write 10=2*5.. Its clear that the no. of higher factors is less or equal to a no. of lower factors[I mean in terms of value].. E.g. in 100=2^2*5^2[Equal no of 5 & 2s] and it has 2 zeros at the end which is merely the no. of 5s.. For 200=2^3*5^2 as it still has 2 fives within it so no. of zeros at the end should be 2.. Check this by taking other examples.. So we can conclude that in base 10 if we do want to get the no. of zeros at the end so it's sufficient to find out no. of 5s within that number... Now check my previous post... If its not clear to you then ask me..I will definately try to help you..Dont hesitate my friend... Check other threads related to the no. of zeros at the end to clear the concept behind of it... Adieu Regards, Dipanjan....
@ dipanjan Thanks now its clear to me but one more doubt is there here on this site only i found one question find the number of zeroes in 15*30*45*................................1500 the solution should be like this 15(1*2*3.....................100) it is 15*!100 so ans will be number of fives in !100 is 24 and 5*3=15 i.e. 25 but in this ans is given 97 how??????? and in solution mentioned that 15^20!100 how???? one more doubt how to find which one is greater in 71^100 and 73^200............
i want some explanation with some example on chienese remainder theorem also.......
Q.Find the no. of zeros at the end of 15*30*45*...*1500 Soln. As I dont know about the link of this soln. so Im trying to xplain it in my way.. Now see in the problem(previously discuused)all no.s are consecutive which form a series..[In the sense of format] But here its not like that.No.s are in a series but with a difference of 15.. And my previous explation holds true for a consecutive kind a series...
Lets take an example.It wud be clear... No. of zeros within 15*30 15*30=2^1*3^2*5^2 As,No. of 5s<No. of 2s So no. of zeros at the end is 1[No. of 2s]
So we have to calculate separetly the no. of 5s and 2s.. 15*30*45*....*1500 =15^100(1*2*3...*100)[Here you did the mistake its not 15*100!] =15^100*100! The power of 5 is 100+24=124 The power of 2 is 97[Calculate the power of 2 in 100!]
As power of 2<power of 5 So no. of zeros at the end is 97....
One thing my friend whenever you will come across a problem which is difficult then straightaway reduce it or cogitate on it by taking easy example...
Btwn 71^100 and 73^200, its very easy to say 73^200 is higher..Base and power both are more for 73^200...
For Chienese Remainder Theorem Go through other threads..There are enough explanation to make it clear..If then also you are in problem then tell it to me... Hope for the other twos above explanation is clear...
Giv ur best to d world Nd d best will come back to u....
sorry but i m not getting how it is 15^100 * !100 please make me clear it and i want to know how to approch this type of problems like which one is greater 888888888888888888^1234567 and 9999999999999999^34567 like this hre i have taken random example . thanks
Actually you are doing a fundamenta mistake... Now see, If m=x*y & n=x*z Then m*n=(x*y)*(x*z)=x^2*y*z
Like the previous one: 15*30=(15*1)*(15*2)=15^2*(1*2)=15^2*2! 15*30*45=(15*1)*(15*2)*(15*3)=15^3*3! ....... 15*30*45*60*....*1500=(15*1)*(15*2)*(15*3)*(15*4)*...*(15*100)[15 has 100 times] =15^100*(1*2*3*...*100) =15^100*100!
-------------------------------------------- What you are calculating that is actually 15+30+45+...+1500 =15(1+2+3+...+100) Which is sum not multiplication.. -------------------------------------------- The answer of 2nd kina qs is not tough.. I am giving a formula It will be handy 2=<(x+1/x)x=<2.8 And try keep in your mind: log 1=0 log 2=.3 log 3=.47 log5=.69 log7=.84
If soln. is not arrived directly or mentally then with the help of Log you can come up with the soln. Another thing in this kind a problem try to make either power or base of that two or more than two digits same...
Cheers. Giv ur best to d world. And d best will cm back to u.....
Soln.:
Summetion of all powers of 5 denotes the no. of zeros at d end...
5^5*10^10*15^15*20^20*25^25...*100^100
5+10+15+20+50(for 25^25)+30+...+200
By visual inspection
Total powers of 5:
(5+10+...+100)(for powers of 5)+(25+50+75+100)(for powers of 25)
=5*20*21/2+25*4*5/2
=1050+250
=1300
So option (b) is correct....
Give your best to the world.
And the best will come back to you....
Regards,
Dipanjan.....
n/a
Hi
can you please give some more insight on this explanation,please explain it in detail.
this funda is not clear to me i have seen some questions on this site like how many zeroes will be in product 15*30*45*.....................1500
it is given there 15^20*!50 i understand !50 but how come 15^20
please explain all this and clear my doubt.
thanks in advance
ans is 1300
Little STar
n/a
Hi
Can you tell me solution in detail and how to approch these type of problems.
Hi guys I am trying to keep this thing understanble..Pls go thru this one..
As we want to find the no. of zeros(in base 10) so we can write 10=2*5..
Its clear that the no. of higher factors is less or equal to a no. of lower factors[I mean in terms of value]..
E.g. in 100=2^2*5^2[Equal no of 5 & 2s] and it has 2 zeros at the end which is merely the no. of 5s..
For 200=2^3*5^2 as it still has 2 fives within it so no. of zeros at the end should be 2..
Check this by taking other examples..
So we can conclude that in base 10 if we do want to get the no. of zeros at the end so it's sufficient to find out no. of 5s within that number...
Now check my previous post...
If its not clear to you then ask me..I will definately try to help you..Dont hesitate my friend...
Check other threads related to the no. of zeros at the end to clear the concept behind of it...
Adieu
Regards,
Dipanjan....
n/a
@ dipanjan
Thanks now its clear to me but one more doubt is there here on this site only i found one question find the number of zeroes in 15*30*45*................................1500
the solution should be like this 15(1*2*3.....................100)
it is 15*!100 so ans will be number of fives in !100 is 24 and 5*3=15 i.e. 25 but in this ans is given 97 how??????? and in solution mentioned that 15^20!100 how????
one more doubt how to find which one is greater in 71^100 and 73^200............
i want some explanation with some example on chienese remainder theorem also.......
Q.Find the no. of zeros at the end of
15*30*45*...*1500
Soln.
As I dont know about the link of this soln. so Im trying to xplain it in my way..
Now see in the problem(previously discuused)all no.s are consecutive which form a series..[In the sense of format]
But here its not like that.No.s are in a series but with a difference of 15..
And my previous explation holds true for a consecutive kind a series...
Lets take an example.It wud be clear...
No. of zeros within 15*30
15*30=2^1*3^2*5^2
As,No. of 5s<No. of 2s So no. of zeros at the end is 1[No. of 2s]
So we have to calculate separetly the no. of 5s and 2s..
15*30*45*....*1500
=15^100(1*2*3...*100)[Here you did the mistake its not 15*100!]
=15^100*100!
The power of 5 is 100+24=124
The power of 2 is 97[Calculate the power of 2 in 100!]
As power of 2<power of 5
So no. of zeros at the end is 97....
One thing my friend whenever you will come across a problem which is difficult then straightaway reduce it or cogitate on it by taking easy example...
Btwn 71^100 and 73^200, its very easy to say 73^200 is higher..Base and power both are more for 73^200...
For Chienese Remainder Theorem Go through other threads..There are enough explanation to make it clear..If then also you are in problem then tell it to me...
Hope for the other twos above explanation is clear...
Giv ur best to d world
Nd d best will come back to u....
Adieu...
Regards,
Dipanjan....
n/a
sorry but i m not getting how it is 15^100 * !100 please make me clear it and i want to know how to approch this type of problems like which one is greater 888888888888888888^1234567 and 9999999999999999^34567 like this hre i have taken random example .
thanks
Actually you are doing a fundamenta mistake...
Now see,
If m=x*y & n=x*z
Then m*n=(x*y)*(x*z)=x^2*y*z
Like the previous one:
15*30=(15*1)*(15*2)=15^2*(1*2)=15^2*2!
15*30*45=(15*1)*(15*2)*(15*3)=15^3*3!
.......
15*30*45*60*....*1500=(15*1)*(15*2)*(15*3)*(15*4)*...*(15*100)[15 has 100 times]
=15^100*(1*2*3*...*100)
=15^100*100!
--------------------------------------------
What you are calculating that is actually
15+30+45+...+1500
=15(1+2+3+...+100)
Which is sum not multiplication..
--------------------------------------------
The answer of 2nd kina qs is not tough..
I am giving a formula It will be handy
2=<(x+1/x)x=<2.8
And try keep in your mind:
log 1=0
log 2=.3
log 3=.47
log5=.69
log7=.84
If soln. is not arrived directly or mentally then with the help of Log you can come up with the soln.
Another thing in this kind a problem try to make either power or base of that two or more than two digits same...
Cheers.
Giv ur best to d world.
And d best will cm back to u.....
Regards,
Dipanjan...........
n/a
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