@mailtoankit @vbhvgupta @VJ12 1300....Let me know how to do it....5 ke powers se hoga but kaise calculate krna hai?
@TootaHuaDil said:1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99 * 100^100.a. 1050b. 1300c. 1250d. None of these.
My take 1300
0 is obtained when you multiply 5 with 2
so there are 5 5's in 5^5(and enough number of 2's)
in 10 there are 10
similary it is summation of 5(1+2+..20)
= 5*20*21/2 = 1050
now in 25^25 you have 25 more
similary in 50^50 and 75^75 and 100^100
so add 25+50+75+100 = 250
so 1050+250 = 1300
@chandrakant.k said:My take 1300
Mast solution....faster :)
Mast solution....faster :)
How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers €Œ? isko kaise krna hai?
@TootaHuaDil said:@mailtoankit@vbhvgupta@VJ12 1300....Let me know how to do it....5 ke powers se hoga but kaise calculate krna hai?
5 + 10 + 15 +.....+100
sum of n terms 5(1+2+3...+ 20) = 1050
and 25 50 75 100 have same no of extra powers
so add them = 250
1050 + 250 = 1300
sum of n terms 5(1+2+3...+ 20) = 1050
and 25 50 75 100 have same no of extra powers
so add them = 250
1050 + 250 = 1300
@TootaHuaDil said:How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers €Œ? isko kaise krna hai?
a+ (a+1 ) ......... (a+8 ) = 9a + 8*9/2 = 9a + 36 9 (a+4 ) = 100 ==> a+4 a= 1 to 7..so,its 7??
@TootaHuaDil said:Mast solution....faster
faster ka matlab??? u mean the method is faster or you want me to be faster ?? i am quite confused 😞
@TootaHuaDil said:How many positive integers less than 100 can be written as the sum of 9 consecutive positive integers €Œ? isko kaise krna hai?
ok here you go.
9*10/2 = 45
now add (10-1) = 9 total = 54
add (11-2) = 9 total = 63
add (12-3) = 9 total = 72
..
go on like this
9*11 tak ayega..
so 11-5 =
plus 1 so total 7 terms can be expressed ??
@TootaHuaDil said:1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99 * 100^100.a. 1050b. 1300c. 1250d. None of these.
OA: Bi.e. 1300
@ravi.theja said:a+ (a+1 ) ......... (a+8 ) = 9a + 8*9/2 = 9a + 36 9 (a+4 ) = 100 ==> a+4 a= 1 to 7..so,its 7??
correcta
@TootaHuaDil numbers of form 9(x + 4) less than 100
where 0 numbers => 9x5, 9x6, .... 9x11 => 7 numbers
@chandrakant.k said:faster ka matlab??? u mean the method is faster or you want me to be faster ?? i am quite confused ok here you go.9*10/2 = 45now add (10-1) = 9 total = 54add (11-2) = 9 total = 63add (12-3) = 9 total = 72..go on like this 9*11 tak ayega..so 11-5 = plus 1 so total 7 terms can be expressed ??
yar didnt get it??
@vbhvgupta said:yar didnt get it??
bhai see this ...will be clear
Do the addition of first 9 natural numbersi.e. n(n+1)/2...here n=9...so sum is 45.
Next take 2-10...exclude 1 from the above sum & add 10..so sum is 54...
next 3-11..exclude 1 nd 2 from 45 nd add 10 nd 11...so sum is 42+10+11=63...
we get a pattern of numbers with a common difference of 9...so less than 100 starting from 45 we have 7 such numbers...45,54,63,72,81,90,99
@vbhvgupta said:yar didnt get it??
arey mai thoda zada hi confusing me bol diya..
see first smmation of 9 numbers will be 1+2+3..9 = 45
now which will be the next number?
you need to add 10 and remove 1 so the difference is 9
now next numbers will be 54 63 72 81 90 99
so what i wrote is 9*5 se lekar 9*11 tak
means 11-5+1 = 7 terms
Hope you undertsand 😃
@TootaHuaDil said:1. What is the number of ending zeros in 1^1 * 2^2 * 3^3........ 99^99 * 100^100.a. 1050b. 1300c. 1250d. None of these.
1300...
Q: In how many ways one can paint a decagon with a blue color.??
