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

#### QUERY 11

What is the unit digit in (795 – 358)?

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 795 = unit digit in 7³ = unit digit in 343 = 3   (We took 7³ in place of 795 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 358 = unit digit in 3² = 3   (We took 3² in place of 358 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 (795 – 358) = 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 = 27
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’)

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