Find the number of zeros at the end of the product 35 x 115 x 40 x 32 x 98 x 22 x 175
Answers
Answer:
1.9438496e+12
You can search for any mathematical expression, using functions such as: sin, cos, sqrt, etc. You can find a complete list of functions here.
(
)
%
AC
7
8
9
÷
4
5
6
×
1
2
3
−
0
.
=
+
Rad
Deg
x!
Inv
sin
ln
π
cos
log
e
tan
√
Ans
EXP
xy
=
123
Fx
the number of zeros at the end of the product 35 x 115 x 40 x 32 x 98 x 22 x 175 is
Step-by-step explanation:
How can you find the number of zeros at the end of expression 15∗32∗25∗40∗75∗98∗112∗125 ?
CIBIL® Score - Now at 10% off | Use promo code SAVE10.
10 is a factor of 5 and 2. In these type of questions you need to know how many 5s and 2s are there in the product. If x is the number of 5s and y the number of 2s, minimum of x and y would be the number of zeroes in any product. In this question
15*32*25*40*75*98*112*125
15 - one 5
32 - five 2s
25 - two 5s
40 - three 2s and one 5
75 - one 3 and two 5s
98 - one 2 and two 7s
112 - four 2s and one 7
125 - three 5s
In total: thirteen 2s and nine 5s.
Minimum of 13 and 9 is 9, which is the number of zeroes in the product.
Save taxes, get guaranteed* returns and a life cover!
See, the number of zeroes at the end of any number/product depend on the number of times 10 occurs in its factorization.
Therefore, to find zeroes at the any product we can simply find 10 in its product.
Since 10=2∗5 , we can further simply our task to finding number of pairs of 2 and 5 in our product.
That being said, our product can be written as 15∗32∗25∗40∗75∗98∗112∗125
=213∗32∗59∗73
We clearly have 13 number of 2 and 9 of 5 .
Given that, we can make maximum 9 pairs of 2 and 5 . This means, we have 9 times 10 in our multiplaction. Which means we will have 9 zeroes at the end of the given product.
In multiplication of numbers ,to find the number of zeroes in the end we just need to find the number of combinations of 5 * 2 .
We know that only 5 * 2 is the only minimum number multiplied to give 0 in the unit place i.e 5 * 2 = 10 (One 0 at unit place)
So,we need to factorize the numbers in the form of 2 and 5.
Eg : 8 = 4 * 2 but we need to write this as 8 = 2 * 2 * 2 because we need the combinations of 5 * 2
Similarly 125 = 25 * 5 but write it as 125 = 5 * 5 * 5 instead
15 = will have one 5 = 3 * 5
32 = five 2
25 = two 5
40 = three 2 and one 5
75 = two 5
98 - one 2
112 - four 2
125 - three 5
Note : I h
Free French class demo. Offer ends tonight!
What is the number of zeros at the end of the product of the number from 1 to 100?
What is number of zeroes at the end of the product 5^5*10^10*15^15*…125^125?
How do I find the number of consecutive zeros at the end of the number 51×102×153×...10020 ?
Prime factorization of each number :
15 = 5 * 3
32 = 2^5
25 = 5^2
40 = 2^3 * 5
75 = 5^2 * 3
98 = 7^2 * 2
112 = 2^4 * 7
125 = 5^3
15*32*25*40*75*98*112*125 = 2^13 * 5^9 * <The Rest>
= (2*5)^9 * 2^4 * <The Rest>
= 10^9 * <The Rest>
Hence 9 zeros :)
It is very easy to find the number of zero at the end ,all you have to do is count how many times did 2 and 5 occured in the question as factor.
Number of zeros is equal to the one (2 or 5)which occured less times.
Eg. If 5 has occurred 7 times and 2 has occured 8 times, then number of zeros if 7.
15=(5*3)
32=(2*2*2*2*2)
25=(5*5)
40=(2*2*2*5)
75=(3*5*5)
98=(7*7*2)
112=(2*2*2*2*7)
125=(5*5*5)
On a Whole there are ((2^13)*(3^2)*(5^9)*(7^3))
We will get zeroes when we multiply 2 and 5
Totally there are(2*5)^9 so there will be 9 zeroes at the end of the equation
Answer is 9
Count the total number of 5*2 groups in the above expression. You will find number of 5 are nine and its frequency is less as compared to 2 so the answer will depend on number of 5.
Number of zeros = total number of 5*2 groups