How many 3 digit numbers are there whose sum of digits is equal to product of digits?

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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Reading Comprehension Vol 1

12 Questions 36 Marks 20 Mins

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.

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2-digit numbers:

Solutions: a = 0, b = 0    &    a = 2, b = 2

22   |—–>   2 + 2   =   4   =   2 * 2

3-digit numbers:

123   |—–>   1 + 2 + 3   =   6   =   1 * 2 * 3
Also true for 132, 213, 231, 312, 321

4-digit numbers:

1124   |—–>   1 + 1 + 2 + 4   =   8   =   1 * 1 * 2 * 4
1142   |—–>   1 + 1 + 4 + 2   =   8   =   1 * 1 * 4 * 2
And other combinations of   1, 1, 2   and   4

5-digit numbers:

11125   |—–>   1 * 1 * 1 * 2 * 5   =   10   =   1 + 1 + 1 + 2 + 5
11152   |—–>   1 * 1 * 1 * 5 * 2   =   10   =   1 + 1 + 1 + 5 + 2
And other combinations of   1, 1, 1, 2   and   5

11133   |—–>   1 * 1 * 1 * 3 * 3   =   9   =   1 + 1 + 1 + 3 + 3
And other combinations of   1, 1, 1, 3   and   3

11222   |—–>   1 * 1 * 2 * 2 * 2   =   8   =   1 + 1 + 2 + 2 + 2
And other combinations of   1, 1, 2, 2   and   2

6-digit numbers:

111126   |—–>   1 * 1 * 1 * 1 * 2 * 6   =   12   =   1 + 1 + 1 + 1 + 2 + 6
111162   |—–>   1 * 1 * 1 * 1 * 6 * 2   =   12   =   1 + 1 + 1 + 1 + 6 + 2
And other combinations of   1, 1, 1, 1, 6   and   2

112411   |—–>   1 * 1 * 2 * 4 * 1 * 1   =   8   =   1 + 1 + 2 + 2 + 1 + 1
And other combinations of   1, 1, 2, 4, 1   and   1

7-digit numbers:

1111127   |—–>   1 * 1 * 1 * 1 * 1 * 2 * 7   =   14   =   1 + 1 + 1 + 1 + 1 + 2 + 7
1111172   |—–>   1 * 1 * 1 * 1 * 1 * 7 * 2   =   14 =   1 + 1 + 1 + 1 + 1 + 7 + 2
And other combinations of   1, 1, 1, 1, 1, 2   and   7

1111134   |—–> 1 * 1 * 1 * 1 * 1 * 3 * 4   =   12 =   1 + 1 + 1 + 1 + 1 + 3 + 4
1111143   |—–> 1 * 1 * 1 * 1 * 1 * 4 * 3   =   12 =   1 + 1 + 1 + 1 + 1 + 4 + 3
And other combinations of   1, 1, 1, 1, 1, 3   and   4
 

8-digit numbers:

11111128   |—–> 1 * 1 * 1 * 1 * 1 * 1 * 2 * 8 =   16   = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 8
11111182   |—–> 1 * 1 * 1 * 1 * 1 * 1 * 2 * 8 =   16   = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 8
And other combinations of   1, 1, 1, 1, 1, 1, 2   and   8
 

9-digit numbers:

111111129   |—–> 1 * 1 * 1 * 1 * 1 * 1 * 1 * 2 * 9 =   18   = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2 + 9
111111192   |—–> 1 * 1 * 1 * 1 * 1 * 1 * 1 * 9 * 2 =   18   = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 9 + 2
And other combinations of   1, 1, 1, 1, 1, 1, 1, 2   and   9

111111135   |—–> 1 * 1 * 1 * 1 * 1 * 1 * 1 * 3 * 5 = 15 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 3 + 5
111111153   |—–> 1 * 1 * 1 * 1 * 1 * 1 * 1 * 5 * 3 = 15 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 5 + 3
And other combinations of   1, 1, 1, 1, 1, 1, 1, 3   and   5

10-digit numbers:

1*1*1*1*1*1*1*1*x*y = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + x + y
x * y = 8 + x + y,   with   0 ≤ x, y ≤ 9
so I get   x = y = 4

1111111144   |—–> 1*1*1*1*1*1*1*1*4*4 =   16   = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 4 + 4
And other combinations of   1, 1, 1, 1, 1, 1, 1, 1, 4   and   4

    Question:   Generalize this.

How many 3 digit numbers are there for which the product of their digits?

What 3 numbers is the sum and product equal?

How many three

How many 3 digit numbers can you find such that product of their digits is a natural number less than 5?