ARITHMETICMATHSNumber, HCF/LCM, Finding of Unit Digit, Sum of Series, etc.

NUMBER OF ZEROES & UNIT DIGIT (PART-I)

NUMBER OF ZEROES & UNIT DIGIT

QUERY 1

To find the number of zeros in a series of number from 1 to 1000.

A) 189
B) 190
C) 192
D) 100

JAYANTCHARAN CHARAN
First note this how to find number of zeroes in a series of number beginning with 1
Note this pattern:

Number of zeros in 1 to 99 = 9
Number of zeros in 1 to 999 = 189
Number of zeros in 1 to 9999= 2889
Number of zeros in 1 to 99999= 38889
Number of zeros in 1 to 999999=488889

And so on………………………..

For example if you want to know the number of zeros in 1 to 100000; we will know the number of zeros in the group 1 to 99999 and will add the number of zeros in 100000 to it.

So the number of zeros in 1 to 100000= 38889+5=38894

NOTE: Remember if we have to find the number of zeros 1 between 100000; not 1 to 100000, the number of zeroes will be 38889 as the number 100000 will not be included in that case.

Therefore numbers of zeros in 1 to 1000 = 189 + 3 = 192 (option ‘C’)

QUERY 2

Find the number of zeros in the product 2*4*6*8*……………….98

A) 42
B) 10
C) 24
D) 32

RONNIE BANSAL
Well, first of all you should understand what FACTORIAL is as this will be the term that is very very helpful in such a problem.
The factorial of a non-integer number n, denoted by n!, is the product of all positive integers less than or equal to n. For example,
5! = 5 x 4 x 3 x 2 x 1

In such a question convert the expression in factorial of a number.
See how: your expression can be written as 249(1 x 2 x 3 x 4……49) = 249(49!)

Now we have to find the number of zeros in the non-factorial part and the factorial part separately and add them. We see the number of zeros in the non-factorial part i.e. 249 is nil.
Now is the turn of factorial part 49! to be considered.

Now follow the following procedure till the divisor does not become greater than the dividend:
X/5 + X/25 + X/125 + X/625 and so on, where X is the value of of the factorial concerned

Here the given number is 49!; so the solution:
49/5 + 49/25 = 9 + 1 = 10 (Remember we have to consider the quotients only while adding)
Hence the answer is 10.

Now take another example. Find the number of zeroes in 1 x 2 x 3 x 4 x 5 x 6 x 7……….98 x 99 x 100.
The expression = 100!
Here we see there is no non-factorial part
Therefore the solution: 100/5 + 100/25 = 20 + 4 = 24
So number of zeroes in the product=24

QUERY 3

The first 100 multiples of 10 are multiplied together. How many zeros will be there at the end of the product?

A) 100
B) 24
C) 124
D) 125

Amit jha
First 100 multiples of 10 = 10, 20, 30, 40…..990, 1000

We have to find out number of zeros in 10 x 20 x 30 x 40 x……x 990 x 1000, which we can write as

10100(1 x 2 x 3 x 4 x…….x 99 x 100)
= 10100(100!)

Now the number of zeros in the non-factorial part i.e. 10100 = 100
And the number of zeroes in the factorial part i.e. 100! = 100/5 + 100/25 = 20 + 4 = 24

So the total umber of zeros in the product = Zeros in non-factorial part + zeros in factorial part i.e. 100 + 24 = 124 (option ‘C’)

QUERY 4

Find the number of digits in 244 × 512

A) 14
B) 12
C) 10
D) 18

JAYANTCHARAN CHARAN
Focus should be on to convert the expression into two such factors that one of them is 10 as base raised to the power of some number & the other factor that can’t have base 10

Now the above expression
= (23 × 3)4 × 512
= (23)4 × 34 × 512
= 212 × 34 × 512
= 212 × 512 × 81
= 1012 × 81
= 10,00,00,00,00,000 × 81
= 8,10,00,00,00,00,000

So the number of digits in the product is 14 (option ‘A’)

QUERY 5

What is the unit place of 122173?

A) 6
B) 3
C) 2
D) 1

MAHA GUPTA
Unit digit in 122173 = unit digit in 2173 = unit digit in 21   (we took 21 in place of 2173 because after every 4 attempts we get the repeated digit when 2 is raised to power of any natural number greater than 4)

So the unit digit in 122173 is  2 (option ‘C’)

QUERY 6

All the even numbers up to 100 are multiplied together. The number of zeros at the end of product?

A) 10
B) 11
C) 12
D) 13

MAHA GUPTA
Well, first of all you should understand what FACTORIAL is as this will be the term that is very helpful in such a problem.
The factorial of a non-integer number n, denoted by n!, is the product of all positive integers less than or equal to n. For example,
5! = 5 x 4 x 3 x 2 x 1

In such a question convert the expression in factorial of a number.
See how: your expression can be written as:
2*4*6*8*10…….98*100; now converting it in factorial 250(1 x 2 x 3 x 4 x 5……50) = 250(50!)

Now we have to find the number of zeros in the non-factorial part and the factorial part separately and add them. We see the number of zeros in the non-factorial part i.e. 250 is nil. (2 raised to any power gives no zero at the end).

Now is the turn of factorial part 50! to be considered.
Now follow the following procedure till the divisor does not become greater than the dividend:
X/5 + X/25 + X/125 + X/625 and so on, where X is the value of of the factorial concerned

Here the given number is 50!; so the solution:
50/5 + 50/25 = 10 + 2 = 12 (Remember we have to consider the quotients only while adding)

Hence the answer is 12 (option ‘C’)

QUERY 7

All the odd numbers up to 100 and 64 are multiplied together. The number of zeros at the end of product?
A) 0
B) 5
C) 6
D) 12

MAHA GUPTA
We have to find the number of zeros at the end of (1 x 3 x 5 x 7 x 9 x 11…………97 x 99) x 64
We see that the above can’t be converted into factorial, hence we need to adopt some different approach.

In such a situation find the number of numbers with 5 at unit’s digit
Obviously they are 5, 15, 25, 35,………………95
They are 10 in all.
We also see out of these 25 and 75 are divisible by 5 twice
Therefore the product of these numbers can be divided by 5 of course 10 + 2 = 12 times.

Now 64 = 26
We know that product of one pair of 2 and 5 i.e. 2 x 5 gives one zero.

Here we have 5 for 12 times, but 2 for 6 times only

Hence total number of zeros at the product = 6 (option ‘C’)

QUERY 8

Find the unit’s digit in 2762n – 1613n, if ‘n’ is a natural number.
A) 1
B) 3
C) 5
D) 6

MAHA GUPTA
Unit digit of 2762n will be the same as will be the unit digit of 62n. Now unit digit of 62n is always 6 for all values of 2n, therefore unit digit of 2762n will also be 6

Unit digit of 1613n will be the same as will be the unit digit of 13n. Now unit digit of 13n is always 1 for all values of 3n, therefore unit digit of 1613n will also be 1

Hence the required answer = 6 – 1 = 5 (option ‘C’)

QUERY 9

The unit’s digit in 71× 63× 653?

A) 1
B) 2
C) 4
D) 5

MAHA GUPTA
When there is multiplication of two or more numbers, we should know the unit digit of each number and multiply those digits; the unit digit of this product will also be the unit digit required.

Unit digit in 717 = unit digit in 17  = 1

Unit digit in 636 = unit digit in 36 = unit digit in 3² = 9 (we took 3² in place of 36 because after every 4 attempts we get the repeated digit when 3 is raised to power of any natural number greater than 4)

Unit digit in 65³ = unit digit in 5³ = 5 (5 raised to power of any natural gives 5 as unit digit)

Now the product of these digits in the manner said above = 1 × 9 × 5 = 45

Hence the required answer is 5 (option ‘D’)

QUERY 10

Find unit digit in 256251 + 36528 + 7354

A) 0
B) 3
C) 4
D) 9

MAHA GUPTA
When there is addition of two or more numbers, we should know the unit digit of each number and add those digits; the unit digit of this addition will also be the unit digit required.

Unit digit in 256251 = 5 (5 raised to power of any natural gives 5 as unit digit)

36528 = 6 (6 raised to power of any natural gives 6 as unit digit)

7354, here we can say that 3² = 9 (we took 3² in place of 354 because after every 4 attempts we get the repeated digit when 3 is raised to power of any natural number greater than 4; the remainder when 54 is divided by 4 = 2)

Now the addition of these digits in the manner said above = 5+6+9 = 20

Hence the required answer is 0 (option ‘A’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
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