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

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Using the digits 0,1,2,...9 how many 7-digit numbers can be formed if the first digit cannot be 0 or 9 and if the last digit is greater than or equal to 2 and less than or equal to 3? Repeated digits are allowed.

Solution

Let the number be 'abc'. Then, $$2<a\times\ b\times\ c<7$$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers.

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers.

Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

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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......

## Answer (Detailed Solution Below) 21

Free

12 Questions 36 Marks 20 Mins

Calculation:

Let the 3-digit number be abc,

⇒ a × b × c = 3 or 4 or 5 or 6

 a × b × c = 3 4 5 6 Possibilities (113, 131, 311) (122, 212, 221) (115, 151, 511) (123, 132, 231) (114, 141, 411) (116, 161, 611) (213, 321, 312)

⇒ Total numbers = 3 + 6 + 3 + 9 = 21

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

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