Maths Puzzle on Number System
Question 1 ) How many 3-digit numbers do not have an even digit or a zero?
__ __ __ - Let this be a 3 digit number.
Since we have to find the number of 3 digit numbers with odd digits (Do not have an even digit or Zero).
Since Zero is an even digit so we have the options of only 1,3,5,7,9 to be filled at these 3 blanks.
For 1st blank number of ways to fill with duplicate allowed is 5 (1,3,5,7,9)
For 2nd blank number of ways to fill with duplicate allowed is 5 (1,3,5,7,9)
For 3rd blank number of ways to fill with duplicate allowed is 5 (1,3,5,7,9)
So,Total numbers of 3 digits in given constrain are
5*5*5=125
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Question 2) How many zeros are in 3 ?
Many zeroes can b added before 3 and after 3 along with a decimal
Like 000000000000000003.0000000000000000
Another solution can be this
There are 3 ‘6’s in 18
There are 2 ‘5’s in 10
There are 10 ‘10’ in 100
Similarly,
There are infinite ‘0′s in 3.
==============================
Question 3 ) How many distinct 3-digit even numbers can be formed?
Since the question is for DISTINCT 3 digit even numbers, the answer can't be 450 is obviously. So below is the answer.
Eg. 124 is distinct 3 digit even number. While 662 is not a distinct 3 digit even numbers.
Now, Since, it is even, last digit can be from 2,4,6,8,0. So total last digits can be 5.
Case 1: when last digit is 2,4,6,8. Suppose when last digit is 2,
Second digit can be from 0 to 9 except 2, to be distinct. Hence total second digits can be 10-1=9.
Since it is to be distinct, Second digit number should not be repeated in first digit (Eg. 662) hence it must be excluded, so first digit number can be from 1 to 9 except last digit and second digit number. Hence total first digit number can be (9-2)=7.
This will be applicable for 4,6,8 also,
So 3 digit number when last digit is (2,4,6,8)=4*9*7=252.
Case 2: when last digit is 0
Second digit can be from 1 to 9 =9.
Again second digit shouldn't be repeated in first digit, Since it has to be distinct.
First digit number can be from (1 to 9 except second digit)=9-1=8.
So 3 digit number when last digit is 0=1*9*8=72.
Hence the answer is 252+72=324
can you answer now for distinct 3-digit odd numbers
MATHS OLYMPIAD PREPARATION BOOK
__ __ __ - Let this be a 3 digit number.
Since we have to find the number of 3 digit numbers with odd digits (Do not have an even digit or Zero).
Since Zero is an even digit so we have the options of only 1,3,5,7,9 to be filled at these 3 blanks.
For 1st blank number of ways to fill with duplicate allowed is 5 (1,3,5,7,9)
For 2nd blank number of ways to fill with duplicate allowed is 5 (1,3,5,7,9)
For 3rd blank number of ways to fill with duplicate allowed is 5 (1,3,5,7,9)
So,Total numbers of 3 digits in given constrain are
5*5*5=125
========================================
Question 2) How many zeros are in 3 ?
Many zeroes can b added before 3 and after 3 along with a decimal
Like 000000000000000003.0000000000000000
Another solution can be this
There are 3 ‘6’s in 18
There are 2 ‘5’s in 10
There are 10 ‘10’ in 100
Similarly,
There are infinite ‘0′s in 3.
==============================
Question 3 ) How many distinct 3-digit even numbers can be formed?
Since the question is for DISTINCT 3 digit even numbers, the answer can't be 450 is obviously. So below is the answer.
Eg. 124 is distinct 3 digit even number. While 662 is not a distinct 3 digit even numbers.
Now, Since, it is even, last digit can be from 2,4,6,8,0. So total last digits can be 5.
Case 1: when last digit is 2,4,6,8. Suppose when last digit is 2,
Second digit can be from 0 to 9 except 2, to be distinct. Hence total second digits can be 10-1=9.
Since it is to be distinct, Second digit number should not be repeated in first digit (Eg. 662) hence it must be excluded, so first digit number can be from 1 to 9 except last digit and second digit number. Hence total first digit number can be (9-2)=7.
This will be applicable for 4,6,8 also,
So 3 digit number when last digit is (2,4,6,8)=4*9*7=252.
Case 2: when last digit is 0
Second digit can be from 1 to 9 =9.
Again second digit shouldn't be repeated in first digit, Since it has to be distinct.
First digit number can be from (1 to 9 except second digit)=9-1=8.
So 3 digit number when last digit is 0=1*9*8=72.
Hence the answer is 252+72=324
can you answer now for distinct 3-digit odd numbers
MATHS OLYMPIAD PREPARATION BOOK
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