Posted on: Thu, 2008-07-24 11:50
please solve it
Hi
let me know the approch to solve this kind of question
1) !N is having 37 zeroes at its end, how many values of N is/are possible
a) 0 b) 1 c) 5 d) infinite
ans c) 5
2) what is the remainder when 22^33+10^35 is divided by 45?
a)2 b) 11 c) 8 d) none of these
ans c) 8
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1) !N is having 37 zeroes at its end, how many values of N is/are possible
a) 0 b) 1 c) 5 d) infinite
ans c) 5
150 have 37 zero becasue 5 comes 37 times
so 150/151/152/153/154 have 37 zero
Little Star
n/a
hi
can you tell me solution in detail hw u approched this how come it came in your mind,i want to know approch to solve this kind of problem please do explain.
the 1st one is simple..
The no. of 'x' in 'N!' is - N/(x^n)+ N/(x^n-1)....+N/(x^1)
Here we hav the no. of 5s in 150 as 37 since 150/5^3+150/5^2+150/5=1(1.2 eqivalent to 1)+6+30=37
but u dont hav to find so much.the next 5 in N! will come after 4 no.s ,i.e. after N+1,N+2..N+4, N+5 will give us another 5 which being multiplied by 2(since no. of 2s r already much greater than 5) will give us another 0 at the end making it 38 zeroes all total.hence there r 5 possibilties-150!,151!,152!,153!,154!.Hope its clear now.
i can't sort out the remainder when 22^33 is divided by 45..10^n will always give 10 as remainder though ,when divided by 45.
n/a
hi
no one is giving me the proper approch i know how to calculate no of x in !N but how u people approched 150 directly that i want to know there is any trick or randomly i want to know that please help me
Hi paridhi,
plz mark the point where u r getting ,so that i can elaborate that particular point further.
Sol: The number of zero is determined by the nos. of 5s & 2s,whichever is less.(since 5 x 2 =10).
In most of the cases number of 5s is less than that of 2s.
This was all about the basics.
Now lets come to the question,
A/q, N/5+N/5^2+N/5^3+N/5^4=37
=> This gives the value of N=150,151,152,153,154
=> 5values.
n/a
@shivanibusiness
can u solve this
1) !N is having 30 zeroes at its end, how many values of N is/are possible
a) 0 b) 1 c) 2 d) infinite
ans c) 0
If we consider N b/w,120 and 125 we get 29 zeroes.
When we consider N b/w, 125 and 130 we get 31 zeroes.
So for no values of N there will be 30 zeroes.
n/a
the numbers whose factorial give 37 zeros at the end are: is 175,176,177,178,179
so the answer is c-5
Explanation:
0s end the end are caused becasue of the following facts:
1. Because of 0s at units place- there are 16 numbers in 1-175 having 0 at units place
2. Because of 00s at LS places- there is one such number that is 100
3. Because of 5s in unitplaces multiplied by even number results in 0 at the end-so such numbers having 5 in units place are 18
4. Because of 5 in 10s place and 0 in unit place-there are 2 such numbers 50 and 150
so total 37 zeros. So 175,176,177,178,179 can form 37 zeros at the end.
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