### CAT 2021 Crash Course + Mock Test Series (INR 4999 Only)

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**Question 1:**

Find the highest power of 30 in 50!

**Answer & Solution**

30 = 2 × 3 × 5. Now 5 is the largest prime factor of 30, therefore, the powers of 5 in 50! will be less than those of 2 and 3. Therefore, there cannot be more 30s than there are 5 in 50! So we find the highest power of 5 in 50! The highest power of 5 in 50! = $\left[ \frac{50}{5} \right]+\left[ \frac{50}{25} \right]$= 10 + 2 = 12. Hence the highest power of 30 in 50! = 12

**Question 2:**

Find the number of zeroes present at the end of 100!

**Answer & Solution**

We get a zero at the end of a number when we multiply that number by 10. So, to calculate the number of zeroes at the end of 100!, we have to find the highest power of 10 present in the number. Since 10 = 2 × 5, we have to find the highest power of 5 in 100! The highest power of 5 in 100! = $\left[ \frac{100}{5} \right]+\left[ \frac{100}{25} \right]$= 20 + 4 = 24

Therefore, the number of zeroes at the end of 100! = 24

**Question 3:**

What is the rightmost non-zero digit in 15!?

**Answer & Solution**

We saw that 15! = 2^{11}× 3^{6}× 5^{3}× 7^{2}× 11 × 13. Now 2^{3}× 5^{3} will give 10^{3} or 3 zeroes at the end. Removing 2^{3}× 5^{3}, we will be left with 2^{8}× 3^{6}× 7^{2}× 11 × 13. Calculating units digit of each prime factor separately, the units digit of the product 2^{8}× 3^{6}× 7^{2}× 11 × 13 = units digit of 6 × 9 × 9 × 1 × 3 = 8.

Therefore, rightmost non-zero digit = 8

**Question 4:**

Find the highest power of 72 in 100!

**Answer & Solution**

72 = 8 × 9. Therefore, we need to find the highest power of 8 and 9 in 72!.

8 = 2^{3}$\Rightarrow$ highest power of 8 in 100! = $\left[ \frac{\left[ \frac{100}{2} \right]+\left[ \frac{100}{4} \right]+\left[ \frac{100}{8} \right]+\left[ \frac{100}{16} \right]+\left[ \frac{100}{32} \right]+\left[ \frac{100}{64} \right]}{3} \right]=32$

9 = 3^{2}$\Rightarrow$ highest power of 9 in 100!= $\left[ \frac{\left[ \frac{100}{3} \right]+\left[ \frac{100}{9} \right]+\left[ \frac{100}{27} \right]+\left[ \frac{100}{81} \right]}{2} \right]=24$

As powers of 9 are less, therefore, powers of 72 in 100! = 24

**Question 5:**

Find the highest power of 24 in 150!

**Answer & Solution**

24 = 8 × 3. Therefore, we need to find the highest power of 8 and 3 in 150!

8 = 2^{3}$\Rightarrow$ highest power of 8 in 150! = $\left[ \frac{\left[ \frac{150}{2} \right]+\left[ \frac{150}{4} \right]+\left[ \frac{150}{8} \right]+\left[ \frac{150}{16} \right]+\left[ \frac{150}{32} \right]+\left[ \frac{150}{64} \right]+\left[ \frac{150}{128} \right]}{3} \right]=48$

Highest power of 3 in 150! = $\left[ \frac{150}{3} \right]+\left[ \frac{150}{9} \right]+\left[ \frac{150}{27} \right]+\left[ \frac{150}{81} \right]=72$

As the powers of 8 are less, powers of 24 in 150! = 48.

**Question 6:**

What is the largest power of 2 that can divide 269!?

**Answer & Solution**

$\left[ \frac { 269 } { 2 } \right] + \left[ \frac { 269 } { 2 ^ { 2 } } \right] + \left[ \frac { 269 } { 2 ^ { 3 } } \right] + \left[ \frac { 269 } { 2 ^ { 4 } } \right] + \left[ \frac { 269 } { 2 ^ { 5 } } \right] + \left[ \frac { 269 } { 2 ^ { 6 } } \right] + \left[ \frac { 269 } { 2 ^ { 7 } } \right] + \left[ \frac { 269 } { 2 ^ { 8 } } \right]$

= 134 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 265

Thus the greatest power of 2 is 265 that can divide exactly 268!

**Question 7:**

How many natural numbers ‘n’ are there, such that ‘n!’ ends with exactly 30 zeroes?

[1] 0

[2] 1

[3] 3

[4] 4

**Answer & Solution**

According to question, n! should have 30 zeroes in the end. If n = 100 (taking randomly keeping in view 30 is the number of zeroes]

100! has $\left\{ \frac { 100 } { 5 } + \frac { 100 } { 5 ^ { 2 } } \right\} = 24$, So n should be greater than 100.

Next multiple of 5 is 105. But 105 = 5 x 21, has only one extra 5. Number of zeroes will increase by 1 only. Similarly 110, 115 and 120 also have one extra 5. Number of zeroes (from 120! to 124!) = 28.

Now, the next multiple of 5 is 125 and 125 contains three 5’s. So, number of zeroes will increase by 3. Number of zeroes in 125! = 28 + 3 = 31. So, there is no factorial of a number which ends with 30 zeroes

**Question 8:**

n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. 1≤n≤1000. How many solutions are possible for ‘n’?

[1] 8

[2] 7

[3] 1

[4] 4

**Answer & Solution**

We can see that increasing the natural number by 1, we are gathering 3 more powers of 5. Therefore, (n + 1) is a multiple of 125 but not a multiple of 625 as it would result in 4 powers of 5. Therefore, (n + 1) will be equal to all the multiples of 125 minus 625.

Total number of multiples of 125 less than 1000 = 8

Total number of multiples of 625 less than 1000 = 1

The required answer is (8 – 1) =7

**Question 9:**

The number 2006! is written in base 22. How many zeroes are there at the end?

[1] 450

[2] 500

[3] 199

[4] 200

**Answer & Solution**

The number of zeroes present at the end of 2006! in base 22 will be equal to the number of times 22 divides 2006! completely. Therefore, we need to find the highest power of 22 contained in 2006!

22 = 2 x 11. As 11 is the largest prime factor of 22. We will find the highest power of 11 contained in 2006!

Therefore, our answer has to be $[ 2006 / 11 ] + \left[ 2006 / 11 ^ { 2 } \right] + \left[ 2006 / 11 ^ { 3 } \right] = 199$