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How many numbers less than \[10,000\] be formed by using different digits from $ 0 $ to $ 9 $ repetition not allowed which are such that they are:
(a) divisible by $ 5 $
(b) divisible by $ 25 $
What will be the corresponding answers for (a) and (b) if repetition is allowed?

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint: First we have to define what the terms we need to solve the problem are.
Since we are going to use the method of permutation to approach this problem which is $ {}^n{p_r} $ ; n is the objects and r is the arrangements ; since the Numbers less than \[10,000\] will at the most contain 4 digits (because in five digits are could not exceed at less than ten thousand) so there are four digits numbers and different digits from $ 0 $ to $ 9 $ is we are going to divide, without repetition.

Complete step by step answer:
(a) divisible by $ 5 $
First one digit number that divisible by five is one; (itself)
Now two digits number that divisible by five is seventeen because those having five in the unit place or zero in the unit place, similarly for the three-digit number we have one thirty-six;
Similarly for the four digits number we have $ 952 $ (four digits that divisible by the five)
Hence, we have total $ 1 + 17 + 136 + 952 = 1106 $ numbers that will divided by five
If the repetition is allowed then there are some changes in the values
First one digit number that divisible by five is one; (itself) Now two digits number that divisible by five is eighteen because those having five in the unit place (excluding zero), similarly for the three-digit number we have one eighty;
Similarly for the four digits number we have $ 1800 $ (four digits that divisible by the five)
Hence, we have total $ 1 + 18 + 180 + 1800 = 1999 $ numbers that will divide by five

(b) divisible by $ 25 $
since the repetition is need not be allowed and will need to divisible by twenty-five
For the first divide number we don’t have any number thus it is zero
for the second digit number we have $ 25,50,75 $ three number;
similarly for the three digits and four digits we count $ 22,154 $ respectively
Hence in total we have $ 0 + 22 + 154 = 179 $
now repetition is allowed and also need to divide by twenty-five;
One digit number is zero;
Two digits number are $ 25,50,75 $
similarly as above; three- and four-digits numbers are $ 36,360 $
Hence in total $ 0 + 3 + 36 + 360 = 399 $

Note: Since less than \[10,000\] means we only need to count up to $ 9999 $
Repetition means that the same number which placed first can be able to repeat again. Also no repetition means that one number cannot be applied twice.