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
Free
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
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.
Last updated on Nov 28, 2022
CAT Exam Slot 3 exam analysis out. The exam was conducted for 198 marks in online mode on 27th November 2022. The notification was released on 31st July 20222. The candidates had applied from 3rd August to 21st September 2022. CAT exam is an entry ticket to get into one of India's top MBA institutes, IIMs, and other premier business schools. The candidates who will qualify in the CAT are eligible to participate in the admissions of MBA institutes
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.