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This is a question asked in India's CAT exam: http://iimcat.blogspot.in/2013/08/number-theory-questions-and-solutions.html

How many numbers with distinct digits are possible product of whose digits is 28?

A. 6

B. 4

C. 8

D. 12

Firstly, I couldn't even understand the question because the English seems grammatically incorrect.

Secondly, I couldn't understand how the answer was arrived at either.

Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.

We cannot have three digits as (2, 2, 7) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are totally 8 numbers.

If you can't have three digits, then how can four digits even be considered? And how on earth did they eventually reach 8 numbers? What does this even mean?

ps: I considered asking on puzzling.stackexchange, but felt that it'd be more appropriate in a math forum.

Community
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Nav
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  • You can have three digits, but then they have to be $1,4,7$, since these are the only divisors of $28$ less than $10$. The solution is correct. They reached eight numbers as follows: The divisors of $28$ are $1,4,7,28$. Obviously $28$ cannot be a digit in base $10$, so our only options are $1,4,7$. So any such number has either two digits or three digits. If it has two, it must be $47$ or $74$. Any of the three permutations of $1,4,7$ give you a digit with this property. There are six such permutations, giving us a total of eight numbrs with this property. – Ethan Alwaise Nov 27 '15 at 04:05
  • Call a natural number $N$ good if all the digits of $N$ are distinct, and the product of the digits of $N$ is $28$. How many good numbers are there? – André Nicolas Nov 27 '15 at 04:06
  • The statement of the problem is incomprehensible. I would award the Exam Board 0/100 for English. – Rob Arthan Nov 27 '15 at 04:07

3 Answers3

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How many numbers with distinct digits are possible product of whose digits is 28?

Best interpretation of that is "How many numbers with distinct digits are possible, when the product of those digits must be 28?"

Well, the prime factorisation of $28$ is $2^2\cdot 7^1$.   That is $28=2\cdot 14 = 4\cdot 7$.   Also, of course, anything multiplied by $1$ is itself.   (Well there's also $2\cdot 2\cdot 7$ but $[2,2,7]$ is not a list of distinct digits. )

So the only lists of distinct digits we can use are $[4, 7]$ and $[1,4,7]$.   There are $2!$ permutations of the former and $3!$ permutations of the later.   Thus we have counted $8$ possibilities.

Graham Kemp
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They arrived at $8$ by counting $2$ possible two-digit numbers (47 and 74) plus $3!=6$ possible three-digit numbers (147, 174, 417, 471, 714, 741). $3!$ counts the number of ways to arrange the three digits $1, 4, 7$ that multiply to $28$.

And you can have three digits, as was demonstrated with 147, etc. That is why they considered four-digit numbers and noted that from that point onward, digits would have to be repeated.

kccu
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Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74.

Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.

We cannot have three digits as (2, 2, 7) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are totally 8 numbers.

Correct Answer: 8

http://iim-cat-questions-answers.2iim.com/quant/number-system/other/number-system_1.shtml

Bharathwaj
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