# 1) How many numbers upto 5 digits can be created using the digits 1,2,3 and 5 each atleast once such that they are multiple of 15?2)A perfect square number having n digits where n is even will have sqaure root with what digit?

BALA SRIVATSAV YERRAMILLI
132 Points
3 years ago
Heyy!

1. For a number to be a multiple of 15, it should be a multiple of 3 and of 5. So, the last digit will be 5 and the sum       of digits should be a multiple of 3. We can have either 4–digit or 5–digit numbers. If we have a 4–digit number,           sum of the digits will be 1 + 2 + 3 + 5 = 11. No 4–digit number formed with digits 1, 2, 3, 5 exactly once can be a       multiple of 3. So, there is no possible 4–digit number.

Now, in any 5-digit number, we will have 1, 2, 3, 5 once and one of these 4 digits repeating once. 1 + 2 + 3 +             5 = 11. So, the digit that repeats for the number to be a multiple of 3 will be 1. In this instance, sum of the                   digits will be 12 and this is the only possibility. So, any 5–digit number will have the digits 1, 1, 2, 3, 5. For the           number to be a multiple of 5, it shall end in 5.
So, the number should be of the form __ __ __ __ 5, with the first 4 slots taken up by 1, 1, 2, 3. These can be            rearranged in 4!2!4!2! = 12 ways. Thus, there are 12 possibilities overall.
1. Since n is even, the square root will contain n/2 digits. For example, if the number is 1225 (n=4), then        the number of digits in the square root = n/2 = 4/2 = 2. And the square root = 35. Thus, the square            root of a perfect square number having n digits will have n/2 digits.
Hope this helps you! :)
BALA SRIVATSAV YERRAMILLI
132 Points
3 years ago
Edited Version ( Sorry for the errors in the previous one)

1. For a number to be a multiple of 15, it should be a multiple of 3 and of 5. So, the last digit will be 5 and the sum       of digits should be a multiple of 3. We can have either 4–digit or 5–digit numbers. If we have a 4–digit number,           sum of the digits will be 1 + 2 + 3 + 5 = 11. No 4–digit number formed with digits 1, 2, 3, 5 exactly once can be a       multiple of 3. So, there is no possible 4–digit number. Now, in any 5-digit number, we will have 1, 2, 3, 5 once and one of these 4 digits repeating once. 1 + 2 + 3 + 5 = 11. So, the digit that repeats for the number to be a multiple of 3 will be 1. In this instance, sum of the digits will be 12 and this is the only possibility. So, any 5–digit number will have the digits 1, 1, 2, 3, 5. For the number to be a multiple of 5, it shall end in 5. So, the number should be of the form __ __ __ __ 5, with the first 4 slots taken up by 1, 1, 2, 3. These can be rearranged in 4!2!4!2! = 12 ways. Thus, there are 12 possibilities overall.

1. Since n is even, the square root will contain n/2 digits. For example, if the number is 1225 (n=4), then        the number of digits in the square root = n/2 = 4/2 = 2. And the square root = 35. Thus, the square            root of a perfect square number having n digits will have n/2 digits.
Hope this helps you! :)
Yashraj
908 Points
3 years ago
1. Practice previous year questions many times..
2. Appear for a test series and analyse it properly (Giving 2–3 hours)
3. Revise your notes multiple times till they enter your soul :)
4. Find your strong points and chapters, work on them..
5. Solve other people’s doubts. It makes your own concepts very clear