Extra Problems for ‘Number of Zeros’ Question Type

Example 1: Find the number of zeros in 2145 x 5234  .
Solution:When we see the question it looks like a very difficult question but this type of question involving number of zeros is very simple and can be solved in seconds. Here the question is asking for the number of zeros, now  we have to look for  the pairs of 2x 5, so the maximum pairs we can form are 145 because we have the  maximum power  of 2 is 145 , so the number of zeros  in this case are 145.

Example 2: How many numbers of zeros are there in following expression
211 x 1253 x 711+ 311 x  711x  210 x  51+  1112 x  213 x 5125
Solution: To explain the solution we must know something about the nature of zeros, before doing this lets have a simple example, the number of zeros in the end of
10+100+1000+……………..+ 10000000000000000000 are 1 because we can take 10 common out of this expression and write it as 10(1+10+100+……………..+ 1000000000000000000) And clearly the term present inside the bracket has unit digit 1 which when multiplied with 10 results into 1 zero at its end.Thus number of zeros in any such expression will depend on the least zero holder in the expression.
Now using the same logic we can say least number of zeros will be contributed by second term i.e. 311 x  711x  210 x  51  as it contains only single power of 5 .Hence it contains only 1 zero, so the number of zeros for whole expression is 1 .

Example 3:Find the number of zeros at end of
5 x 10 x 15 x 20 x 25 x 30 x 35…………………………………… x 240 x 245 x 250
Solution: We can take 5 common out of this expression and write it as
550 (1 x 2 x 3 x 4……………………………………  x 49 x 50)
550 (50!)
To find number of zeros we first find
Maximum power of 2 in 50!

$\displaystyle \begin{array}{l}\left[ \frac{50}{2} \right]+\left[ \frac{50}{{{2}^{2}}} \right]+\left[ \frac{50}{{{2}^{3}}} \right]+\left[ \frac{50}{{{2}^{4}}} \right]+\left[ \frac{50}{{{2}^{5}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }25\text{ }+\text{ }12\text{ }+\text{ }6\text{ }+\text{ }3\text{ }+1 \\ =\text{ }47 \\ Maximum\text{ }power\text{ }of\text{ }5\text{ }in\text{ }50! \\ \end{array}\\\left[ \frac{50}{5} \right]+\left[ \frac{50}{{{5}^{2}}} \right]+\left[ \frac{50}{{{5}^{3}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }10\text{ }+\text{ }2 \\ =\text{ }12 \\ \end{array}\end{array}$

Thus maximum power of 5 present in given expression is 62 and maximum power of 2 present in given expression is 47. Hence number of zeros will be 47.

Assignment:
Questions:
1 : Find the number of zeros in 75!
a) 16
b) 18
c) 20
d) 21
Ans: b
Solution:
Maximum power of 5 in 75!
$\displaystyle \begin{array}{l}\left[ \frac{75}{5} \right]+\left[ \frac{75}{{{5}^{2}}} \right]+\left[ \frac{75}{{{5}^{3}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }15\text{ }+\text{ }3 \\ =\text{ }18 \\ Hence\text{ }number\text{ }of\text{ }zeros\text{ }will\text{ }be\text{ }18. \\ \end{array}\end{array}$

2: Find the number of zeros in 255!
a) 63
b) 52
c) 62
d) 65
Ans: a
Solution:
Maximum power of 5 in 255!
$\displaystyle \begin{array}{l}\left[ \frac{255}{5} \right]+\left[ \frac{255}{{{5}^{2}}} \right]+\left[ \frac{255}{{{5}^{3}}} \right]+\left[ \frac{255}{{{5}^{4}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }51\text{ }+\text{ }10\text{ }+\text{ }2 \\ =\text{ }63 \\ Hence\text{ }number\text{ }of\text{ }zeros\text{ }will\text{ }be\text{ }63. \\ \end{array}\end{array}$

3: Find the number of zeros in 135!

a) 25
b) 30
c) 33
d) 32
Ans: c
Solution:
Maximum power of 5 in 135!
= 27 + 5 + 1
= 33
$\displaystyle \begin{array}{l}\left[ \frac{135}{5} \right]+\left[ \frac{135}{{{5}^{2}}} \right]+\left[ \frac{135}{{{5}^{3}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }27\text{ }+\text{ }5\text{ }+\text{ }1 \\ \begin{array}{l}=\text{ }33\\Hence\text{ }number\text{ }of\text{ }zeros\text{ }will\text{ }be\text{ }33.\end{array} \\ \end{array}\end{array}$

4: Find the number of zeros in n! where n is number between 66 to 69
a) 12
b) 14
c) 15
d) 16
Ans: c
We pick any value of n between 66 and 69. Number of zeros will be same for any value we pick between 66 and 69 say 68
Maximum power of 5 in 68!
= 13 + 2
= 15

$\displaystyle \begin{array}{l}\left[ \frac{68}{5} \right]+\left[ \frac{68}{{{5}^{2}}} \right]+\left[ \frac{68}{{{5}^{3}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }13\text{ }+\text{ }2 \\ =\text{ }15 \\ \end{array}\end{array}$
Hence number of zeros will be 15.

5: Find the number of zeros in 350!

a) 84
b) 85
c) 86
d) 87
Ans: c
Solution:
Maximum power of 5 in 350!
= 70 + 14 + 2
= 86

$\displaystyle \begin{array}{l}\left[ \frac{350}{5} \right]+\left[ \frac{350}{{{5}^{2}}} \right]+\left[ \frac{350}{{{5}^{3}}} \right]+\left[ \frac{350}{{{5}^{4}}} \right]+…..\\\begin{array}{*{35}{l}} =\text{ }70\text{ }+\text{ }14\text{ }+\text{ }2 \\ =\text{ }86 \\ \end{array}\end{array}$
So the number of zeros in the end of the 350! are 86.