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The product of the positive integers 1 to 6 is:
6 x 5 x 4 x 3 x 2 x 1 = 720
and can be written in an abbreviated form as 6!. We say, "6 factorial ". So 6! = 720.
The product of the positive integers from 1 to 12 is
12 x 11 x 10 ... 3 x 2 x 1 = 479 001 600
and can be written in an abbreviated form as 12!. We say, "12 factorial ".
The ... represents the product of all of the missing integers between 10 and 3.
For a positive integer n, the product of the positive integers from 1 to n is n!.
Find the smallest possible value of n such that n! ends in exactly six zeroes.
Hi. My sister and I worked on this question together.
ReplyDeleteWe did trial and error to find our answer to be 18!.
Based on what the question is asking us to find, we know that we have to find the value of n so that the product ends with 6 zeroes. From here, we tried to find the factorial of every positive number after 1 (2,3,4...8,9,10 etc.) With this method, the first number that ends with 6 zeroes must be the smallest possible value of n. We stumbled upon the number 18. 18! is 6.402373706x10^15. We then converted this number to standard form, which is 64,032,737,706,000,000. Since 18! fits the description (ends with 6 zeroes), it is correct.
- Jonathan and Rachel
Good job Rachel and Jonathan!
DeleteHowever, I don't get how 18 has 6 zeroes.
Hello
DeleteYou guys did a very good job finding the answer with a good process. But can you explain to me how you converted 6.402373706x10^15 into standard form pls
Thanks
Hiya
18!
ReplyDelete2x5
ReplyDelete10
12x15
20
22x25
all end in 0.
Since 25=5^2,
this adds another 0.
Therefore, the number 25! ends in 6 zeroes.
To find the answer I used only elimination, trial and error to figure out the answer. First I began by finding the products of the factorials 12 and above because above the Mr.Millete's explanation tells me that numbers below 12! won't work. So I began by checking numbers above 12.
ReplyDelete13!=6,227,020,800 (wrong)
14!=87.178,291,200 (wrong)
15!=1,307,674,368,000 (wrong)
16!=20,922,789,888,000 (wrong)
17!=355,687,428,096,000 (wrong)
18!= 6,402,373,705,728,000,000 (correct= 6.4023737e+15)
Therefore I came upon 18!= 6,402,373,705,728,000,000 with 6 zeroes on the end and the lowest values.
Careful with 18! This is actually not going to give us the six zeroes we need. Please refer the George_Lizard's answer or below:
ReplyDeleteWe observe that 5! ends in 0 and 10! ends in 00. Notice that the number of
zeros at the end of the number increased by one at each of 5! and at 10!. Why
is this?
A zero is added to the end of a number when we multiply by 10. Multiplying a
number by 10 is the same as multiplying a number by 2 and then by 5, or by 5
and then by 2, since 2 x 5 = 10 and 5 x 2 = 10. We must determine the next
time we multiply by 2 and 5 (in some order), to know the next time the
number of zeros at the end of the number increases again. Every time we
multiply by an even number we are multiplying by at least one more 2. There
are less multiples of ve. Each multiple of ve will a ect the number of zeroes
at the end of the product.
Multiplying by 11, 12, 13, and 14 increases the number of 2's we multiply by
but not the number of 5's. So the number of zeros at the end of the product
does not change. The next time we multiply by a 5 is when we multiply by 15
since 15 = 5 x 3. So 15! will end in exactly three zeroes, 000.
Multiplying by 16, 17, 18, and 19 increases the number of 2's we multiply by
but not the number of 5's. So the number of zeros at the end of the product
does not change. The next time we multiply by a 5 is when we multiply by 20
since 20 = 4 x 5. So 20! will end in exactly four zeroes, 0000.
Multiplying by 21, 22, 23, and 24 increases the number of 2's we multiply by
but not the number of 5's. The next time we multiply by a 5 is when we
multiply by 25. In fact, multiplying by 25 is the same as multiplying by 5 twice
since 25 = 5 x 5. So when we multiply by 25, we will increase the number of
zeros on the end of the product by two. So 25! will end in exactly six zeroes,
000 000.
Then, for n! to end in six zeroes, the smallest value of n is 25. That is, 25! is
the smallest factorial that ends in exactly six zeroes. (It could be noted that
26!; 27!; 28!; and 29! also end in six zeroes.)