**NUMBER OF ZEROES IN AN EXPRESSION****Step 1 :**Suppose you have to find the number of zeroes in a product: 24 × 32 × 17 × 23 × 19. We first get the series in terms of its prime factors i.e. \( 2^3 * 3^1 * 2^5 * 17^1 * 19 * 23\). As you can notice, this product will have no zeroes because it has no 5 in it.However, if you have an expression like: 8 × 15 × 23 × 17 × 25 × 22 The above expression can be rewritten in the standard form as: \( 2^3 * 3^1 * 5^1 * 23 * 17 * 5^2 * 2^1 * 11^1 \)

**Step 2 :**Zeroes are formed by a combination of 2 × 5. Hence, the number of zeroes will depend on the number of pairs of 2’s and 5’s that can be formed. In the above product, there are four twos and three fives. Hence, we shall be able to form only three pairs of (2 × 5). Hence, there will be 3 zeroes in the product.**Finding the Number of Zeroes in a Factorial Value :**Suppose you had to find the number of zeroes in 6!. 6! = 6 × 5 × 4 × 3 × 2 × 1 = (3 × 2) × (5) × (2 × 2) × (3) × (2) × (1). Counting the number of 5 will give the answer.

**Method 2 :**For finding the zeroes in 6! we use \( \frac{6}{5} + \frac{6}{5^3} + ... \) So we get 1 as the answer as all divisions after the first term in the series are in decimals which we ignore.

**Ans . **

10

**Explanation :**\( 47! = \frac{47}{5} + \frac{47}{5^2} + \frac{47}{5^3} + ... \) = 9 + 1 + 0 = 10

**Ans . **

13

**Explanation :**\( \frac{58}{5} + \frac{58}{5^2} + ... \) = 11 + 2 + 0 = 13

**Ans . **

3

**Explanation :**\( 13 * 3 * 5 * 11 * 2 * 5^3 * 11 * 2^2 * 5 * 7 * 11 \) = pairs of (5 * 2) are three so zeroes are 3.

**Ans . **

2

**Explanation :**\( 2^2 * 3 * 3 * 5 * 5 * 2^3 * 3 * 13 * 17 \) = so pairs of (5*2) are 2, so we have 2 zeroes.

**Ans . **

41

**Explanation :**\( \frac{173}{5} + \frac{173}{5^2} + \frac{173}{5^3} = 34 + 6 + 1 = 41 \)

**Ans . **

37

**Explanation :**\( \frac{144}{5} + \frac{144}{25} + \frac{144}{125} + 5 * 3 * 5 * 11 * 2 * 2^2 * 11 * 5 * 3^3 = 28 + 5 + 1 + 3 = 37\)

**Ans . **

35

**Explanation :**\( \frac{148}{5} + \frac{148}{5^2} + \frac{148}{5^3} = 29 + 5 + 1 = 35\)

**Ans . **

370

**Explanation :**\( \frac{1142}{5} + \frac{1142}{25} + \frac{1142}{125} + \frac{1142}{625} + \frac{348}{5} + \frac{348}{5^2} + \frac{348}{5^3} + \frac{17}{5} = 228 + 45 + 9 + 1 + 69 + 13 + 2 + 3 = 370\)

While solving 45!, 46!, 47!, 48!, 49!. Notice the number of zeroes in each of the cases will be equal to 10.

It is not difficult to understand that the number of fives in any of these factorials is equal to 10. The number of zeroes will only change at 50! (It will become 12).

In fact, this will be true for all factorial values between two consecutive products of 5.

Thus, 50!, 51!, 52!, 53! And 54! will have 12 zeroes (since they all have 12 fives). Similarly, 55!, 56!, 57!, 58! And 59! will each have 13 zeroes.

While there are 10 zeroes in 49! there are directly 12 zeroes in 50!. This means that there is no value of a factorial which will give 11 zeroes. This occurs because to get 50! we multiply the value of 49! by 50. When you do so, the result is that we introduce two 5’s in the product. Hence, the number of zeroes jumps by two.

**Note :**at 124! you will get 24 + 4 fi 28 zeroes. At 125! you will get 25 + 5 + 1 = 31 zeroes. (A jump of 3 zeroes.)

**Ans . **

none

**Explanation :**This can never happen because at 99! number of zeroes is 22 and at 100! the number of zeroes is 24.

**Ans . **

59 and 55

**Explanation :**At 55 we get 13 zeroes, since we know that 50! is 12 zeroes so till 54! we will have 12 zeroes. So 55 to 59! will have 13 zeroes.

**Ans . **

250

**Explanation :**The fives will be less than the twos. Hence, we need to count only the fives. \( 5^5 + 10^10 + 15^15 + ... + 45^45 = 5 + 10 + 15 + 20 + 25 + 30 + 35 +40 + 45 = 250 \)

**Ans . **

1124

**Explanation :**Again the key here is to count the number of fives. This can get done by: \( 100^1 × 95^6 × 90^{11} × 85^{16} × 80^{21} × 75^{26} × …… 5^{96} \) = (1 + 6 + 11 + 16 + 21 + 26 + 31 + 36 + 41 + 46 + …….. + 96) + (1 + 26 + 51 + 76) = 20 × 48.5 + 4 × 38.5 = 970 + 154 = 1124.

**Ans . **

5! + 10!

**Explanation :**The answer will be the number of 5’s. Hence, it will be 5! + 10!

**Ans . **

98

**Explanation :**The number of fives is again lesser than the number of twos. The number of 5’s will be given by the power of 5 in the product: 5

^{4}× 10^{8}× 15^{12}× 20^{16}× 10^{18}× 25^{20}= 4 + 8 + 12 + 16 + 18 + 40 = 98.

**Ans . **

**Explanation :**Sum is 15152400, hence two zeros

**Ans . **

18

**Explanation :**The number of zeroes will be: 2 + 3 + 4 + 3 + 6 = 18.

**Ans . **

12

**Explanation :**\( \frac{81}{7} + \frac{81}{7^2} + \frac{81}{7^3} = 11 + 1 = 12\)

**Ans . **

19

**Explanation :**42 = 2 *3 * 7. Hence we must find number of occurances of 2,3,7 in 122!. It is obvios that 7 will have the least occurances. Hence \( \frac{122}{7} + \frac{122}{7^2} + \frac{122}{7^3} = 17 + 2 + 0 = 19 \)

**Ans . **

128

**Explanation :**360 = 5 × 2 × 2 × 2 × 3 × 3. This means we need to think of which amongst 2

^{3}, 3^{2}and 5 would appear the least number of times in 520! Number of 5’s in 520!Highest power of 5 in 520! = 104 + 20 + 4 = 128. In order to find the number of 3^{2}s in 520! we first need to find the number of 3’s in 520! Number of 3’s in 520! = 173 + 57 + 19 + 6 + 2 = 257. 257 threes would obviously mean [257/2]= 128 -> 2^{2}s. In order to find the number of 2^{3}s in 520! we first need to find the number of 2’s in 520! Number of 2’s in 520! = 260 + 130 +65 + 32 +16 + 8 + 4 + 2 + 1 = 518. 518 twos would obviously mean [518/3]= 172 2^{3}s. Thus, the highest power of 360 that would divide 520! would be the least of 128, 128 and 172. The answer is 128.

**Ans . **

38

**Explanation :**Question is same as asking to find number of zeroes in the factorial. This is found by [157/5] = 31. [31/5] = 6. [6/5] = 1. 31 + 6 + 1 = 38. Option (b) is correct.