# NUMBER OF ZEROES & UNIT DIGIT (PART-II)

#### QUERY 11

**What is the unit digit in (7 ^{95} – 3^{58})?**

A) 0

B) 4

C) 6

D) 7

**MAHA GUPTA
**When there is subtraction of two numbers, we should know the unit digit of each number and subtract the second digit from the first digit.

Unit digit in 7^{95} = unit digit in 7³ = unit digit in 343 = 3 (We took 7³ in place of 7^{95} because after every 4 attempts we get repeated digits when 7 is raised to power of any natural number greater than 4.)

Unit digit in 3^{58} = unit digit in 3² = 3 (We took 3² in place of 3^{58} because after every 4 attempts we get repeated digits when 3 is raised to power of any natural number greater than 4.)

Therefore, Unit digit in (7^{95} – 3^{58}) = 3 – 9 = 4 (When the first number is smaller than the second in a question on finding unit digit in case of subtraction we add 10 to that and then subtract.) [option ‘B’)

#### QUERY 12

**If x and y are positive integers such that (3x + 7y) is a multiple of 11, then which of the following will be divisible by 11?**

A) 4x + 6y

B) x + y + 4

C) 9x + 4y

D) 4x – 9y

**MAHA GUPTA
**In such a question one should use HIT AND TRIAL METHOD, otherwise it could be very lengthy to do.

So, by hit and trial, we put x = 5 and y = 1 so that (3x + 7y) = (3*5 + 7*1) = 22, which is a multiple of 11 of course.

Now let’s check the answer options

4x + 6y = 4*5 + 6*1 = 26, is not divisible by 11

x + y + 4 = 5 + 1 + 4 = 10, is not divisible by 11

9x + 4y = 9*5 + 4*1 = 49, is not divisible by 11

4x – 9y = 4*5 – 9+1 = 11, is divisible by 11, hence the answer (option ‘D’)

#### QUERY 13

**The numbers 1, 3, 5, 7, …, 99 and 128 are multiplied together. The numbers of zeros at the end of the product must be?**

A) 19

B) 22

C) 7

D) Nil

**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 128

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 128 = 2^{7}

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 7 times only

Hence total number of zeros at the product = 7 (option ‘C’)