How many 3 digit numbers are there for which the product of their digits is more than 2 but less than 7 ?\?

The question is from Permutation and Combination. Another question which combines number theory with combinatorics. Our task is to find number of 3 digits numbers which satisfies the given condition. This section hosts a number of questions which are on par with CAT questions in difficulty on CAT Permutation and Combination, and CAT Probability.

Question 24: Find 3 digit numbers such that product of their digits is a natural number less than 5?

1. 11
2. 15
3. 13
4. 17

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Explanatory Answer

Method of solving this CAT Question from Permutation and Combination: 3 digit number such that the product of their digits is less than 5

For the product of a 3 digit number to be lesser than 5, it should comprise of either 1, 2 ,3 or 4 in a certain way.
Let’s start with product = 1 => this is possible in only one way (111).
Product =2 => Possible in 3 ways(211, 112 and 121).
Product = 3=> Possible in 3 ways(311, 113 and 131).
Product = 4 => Possible in 3 ways(411, 114, 141, 221, 212 and 122).
Total = 1+3+3+6 =13 numbers.

The question is "Find 3 digit numbers such that product of their digits is a natural number less than 5?"

Hence the answer is "13"

Choice C is the correct answer.

Slot – 1 – Quantitative Aptitude – Number Series – How many 3-digit numbers are there

Q. How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?
Answer: 21

Solutions:
Let the 3-digit number be abc so possible values of a×b×c are 3,4,5 and 6.
If a×b×c=3,(a,b,c)=(1,1,3) ,total such numbers=3
If a×b×c=4,(a,b,c)=(1,1,4) or (1,2,2),total such numbers=6
If a×b×c=5,(a,b,c)=(1,1,5) ,total such numbers=3
If a×b×c=6,(a,b,c)=(1,1,6),(1,2,3) ,total such numbers=9
Thus total possible values of abc = 3+6+3+9 = 21

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Hint:let the number be $\overline{xyz}$

Now if $x\cdot y\cdot z=p$ where $p=3,5$ then we $x,y,z$ must be a permutation of $1,1,p$ Hence number of such numbers is $2\cdot \frac{3!}{2!}$

Now if $x\cdot y\cdot z=4,6$ then $x,y,z$ must be a permutation of either $1,1,4$ or $2,2,1$ or $1,1,6$ or $2,3,1$ .Each of which the number of cases can be easily found......

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Option: 21
Solution:

The product of the digits of the three-digit numbers should be more than 2 and less than 7 . Hence the possible numbers are as follows.

Hence there are a total of 21 possibilities.

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How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Answer (Detailed Solution Below) 21

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Calculation:

Let the 3-digit number be abc,

⇒ a × b × c = 3 or 4 or 5 or 6

 ⇒ Total numbers = 3 + 6 + 3 + 9 = 21

∴ There are 21 digits whose products of their digits is more than 2 but less than 7.

Last updated on Jan 1, 0001

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