Question

How many natural numbers are there which are smaller than 10000 and are divisible by 4, whose decimal notation consists only of the digits 0,1,2,3 and 5 which do not repeat in any of these numbers?

Answer 1

Hint: Try to find the total number of ways to get 1-digit, 2-digit, 3-digit, 4-digit, 5-digit numbers separately then add them and then subtract the total number of ways of getting a 1 digit number. You will get the answer.
Complete Step by Step Solutions:
For 1-digit;
There are a total 5 numbers here so we can choose each of them once to get a 1 digit number i.e., 0,1,2,3,5. Therefore the total number of ways are 5.
For 2-digit;
The tens place cannot be a 0 therefore for tens place we have 4 options and for ones place 0 can be there therefore for ones place we can have 4 different options without repetition of digits
Therefore total number of ways are \[4 \times 4 = 16\]
For 3-digit;
Now the hundredth place cannot be 0 but we can have 0 in tens and ones places.Therefore we have a total of 4 option for hundreds place, 4 option for tens and 3 options for ones that makes it
\[4 \times 4 \times 3 = 48\]
For 4-digit;
Similarly we cannot have a 0 in the thousands place but it can be present in hundred tens and ones. So now we have 4 options for thousand places, 4 options for hundreds places, 3 options for tens and 2 options for one place. So we are getting \[4 \times 4 \times 3 \times 2 = 96\]
Therefore total numbers that will be less than 10000 and formed by 0,1,2,3 and 5 are
\[5 + 16 + 48 + 96 = 165\]
Now the total numbers that will be divisible by 4 can be obtained just by subtracting the single digits number, for 2, 3 and 4 digits number we have to check weather the last two numbers are divisible by 4 or not Clearly the last digits has to be either 0 or 2 for the number to be an even number. Now in 2 digit numbers if we fix the last digit as 0 the only possibility for the first digit is to be 2 because 10, 30, 50 are not divisible by 4. Now if we put the last digit number to be 2 the first place can have 1, 3 and 5 because 12, 32 and 52 all are divisible by 4. So for 2 digit number we have 4 possibilities i.e., 20, 12, 32 and 52
Now for the 3 digit number we know that the last 2 digits has to be divisible by 4 so clearly the last 2 digits has to be one of the 2-digits numbers. Therefore for 20 we have 3 possibilities the first digit can be 1, 3 or 5. Similarly for the rest 3 numbers we have 2 possibilities each as the first number cannot be zero. Therefore for 3-digit number we have \[3 + 2 + 2 + 2 = 9\]
Moving on to 4-digit numbers, again the last 2 digits has to be either of the 2-digit numbers.
Therefore, For 20 we can choose any 2 from the remaining 3-digits and that can be done as we can choose any one from the 3 for the first place and any one of the two for the second place Therefore for the last digits to be 20 we have \[3 \times 2 = 6\] ways. For the remaining 3 2-digit numbers we can choose one from 2 for the first place as the first place cannot be zero and for the second digit zero can be there therefore it also has 2 choices so all of them will have \[2 \times 2 = 4\] ways to form. So the total 4-digit numbers are \[6 + 4 + 4 + 4 = 18\]
So we have 0 numbers n 1-digit, 4 from 2-digit, 9 from 3-digit and 18 from 4-digit numbers
Therefore the total numbers are \[0 + 4 + 9 + 18 = 31\]
Therefore 31 is the correct answer.

Note: We haven't counted 5-digit numbers because 10000 is the smallest 5-digit number and it is also specified that the number has to be less than 10000, also note that for 2 we only check the last digit of the number and say whether it will be divisible by 2 or not. We check last 2 digits for 4 and last 3 digits for 8 because \[{2^1} = 2,{2^2} = 4,{2^3} = 8........\] similarly the total power of 2 it will have we will check those many digits from the last.