Oodles of 0's

[mxs2009]

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

I calculated this in excel and the answer came out to 67 0's at the end. what am I missing?

[susan.meng]

Another way to write 60! = k *10^p where k is some constant

For a number to end in 0, it must be divisible by 10
10^p = 2^p * 5^p
We are looking for the number of 5's in 60!.

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

There are 14 5's (two in 25 and 2 in 50 and 1 in each of the other numbers).

Answer is C.

[mxs2009]

Thanks. I realized that excel rounds the calculation

[deepak1982]

This can also be done by

Successively divide 60 by 5 till u get a quotient which is less than 5.

24/5 = 12
12/5 = 2

So the total no.of Zeros = 12+2 = 14

[mxs2009]

I don't think that shortcut works. try it with 30!. the answer is 7 and your method results in an answer of six.

[RonPurewal]

hi -

just checking - do you have a question about the actual problem or its solution, or are you just wondering about the excel discrepancy?

thanks

[girl.jobless]

The value of 60! = 8.32098711 × 10^81 as per google. So that would mean that there are 73 0's following the integer 832098711. So how is 14 as the answer correct?

[vishalsahdev03]

I calculated 14 0s, i wonder what is google saying.
Answer is C.
Any reviews on this?

[RonPurewal]

I calculated this in excel and the answer came out to 67 0's at the end. what am I missing?

The value of 60! = 8.32098711 × 10^81 as per google.

whoa people. these are rounded values; google and excel are not going to give you ALL the digits of such a huge number.

first, to write all eighty digits of such a number on the screen would be, to say the least, unsightly.
second, it would be terrible for comprehension.
(quick, which is bigger: 48372938478394095827393827840293 or 443222334688970552213733257840677? now, which is bigger: 3.83 x 10^21, or 2.63 x 10^22?)

if you want proof that the values are rounded, type in something like 5^90 into either google or excel.
ALL exact powers of 5 end with 5, but your answer most likely won't. this indicates rounding.

[RonPurewal]

If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A. 6
B. 12
C. 14
D. 42
E. 56

I calculated this in excel and the answer came out to 67 0's at the end. what am I missing?

exact solution:
each "0" comes from the presence of a 10 in the number; each 10 will come from a 2 and a 5.
the limiting factor is whichever of these are fewer. it should be clear that there will be WAY more 2's than 5's, so we only have to figure how many 5's are in the product.
you get ONE "5" each from the following numbers:
5, 10, 15, 20, 30, 35, 40, 45, 55, 60
you get TWO "5"s from:
25, 50
so that's fourteen 5's total.

[dul_hadita]

you get ONE "5" each from the following numbers:
5, 10, 15, 20, 30, 35, 40, 45, 55, 60 there are 10 5s here
you get TWO "5"s from:
25, 50 additional 2 s's here
so that's fourteen 5's total.

that is 12 5's in total.

2x5, 10,12x15, 20, 22x25, 30, 32x35, 40, 42x45, 50, 52x55, 60

25 and 50 will each get you an adiitional 5. so 10+2=12

[RonPurewal]

that is 12 5's in total.

nope -- see your own work:

2x5, 10,12x15, 20, 22x25, 30, 32x35, 40, 42x45, 50, 52x55, 60

this is already twelve 5's.
i'm not sure why you are introducing all of the extra multiples of 2 -- we know that we have an excess of 2's, so we really don't have to bother thinking about multiples of 2.

25 and 50 will each get you an adiitional 5. so 10+2=12

...so the addition of these two extra 5's brings the total to fourteen.

[mirzank]

I'm going to re open an old questions, i got this question wrong and just wanted to show my methodology. Maybe I missed something in my method, or maybe tottaly wrong if someone can enlighten me would appreciate it.

The way I looked at it was to say that the 0's in 60! will either come from the multiples of 10, i.e. 10, 20, 30, 40, 50, 60, which is 6 0's.

Plus another set of zero's will come from 2 and 5 being multipled from units digits. This way i get another 6 0's. 2 and 5, 12 and 15, 22 and 25, 32 and 35, 42 and 45, & 52 and 55.

This way I only get 12 0's. Is this totally the wrong way to think about this? If not, then what can i add to this method to get the 2 extra 0's.

that is 12 5's in total.

nope -- see your own work:

2x5, 10,12x15, 20, 22x25, 30, 32x35, 40, 42x45, 50, 52x55, 60

this is already twelve 5's.
i'm not sure why you are introducing all of the extra multiples of 2 -- we know that we have an excess of 2's, so we really don't have to bother thinking about multiples of 2.

25 and 50 will each get you an adiitional 5. so 10+2=12

...so the addition of these two extra 5's brings the total to fourteen.

[tim]

You missed the same two 0’s that have been discussed extensively in this thread - the extra 0’s that derive from 25 and 50. Please re-read the discussion above so you can see where the extra two 0’s come from ..

[iconstudent475]

Sorry for reopening this, but Excel give me this number when i calculate 60! :
8320987112741390000000000000000000000000000000000000000000000000000000000000000000.00

Much more than 14 zeros...

Did i miss anything ?
Pierre