Question

How many two-digit numbers with distinct digits can be formed such that the product of the digits is the cube of a positive integer?

Select one:
a. 4
b. 3
c. 2
d. 5

Answer 1

Let the two digit number be xy where x is the units digit and y is the ones digit.

Number = 10x + y (x ≠ y)

If we calculate the maximum possible cube of an integer with this technique, we get that we can get xy = 9 x 8 = 72 at the most.

Number of integral cubes till 72 = 5.

They are,

0^{3} = 0

1^{3} = 1

2^{3} = 8

3^{3} = 27

4^{3} = 64

Notice that no more integer is possible which has a higher value than 4

As 5^{3} = 125 which is greater than our limit of 72.

0 can be formed by using any other number from 1 to 9.

They are - 10, 20, 30, 40, 50, 60, 70, 80 and 90

All these numbers satisfy the conditions that are given in the question. They account for 9 numbers.

[Note that we could not use 01, 02 etc. as they are one digit numbers]

Then, we notice that no two distinct integers, when multiplied give product as 1.

Therefore, we exclude the possibility of getting 1 as that product with such number.

Then we see that 2^{3} = 8 can be obtained from these following numbers :

18, 81, 42 and 24. These account for further 4 numbers. So till now, we have got 13 numbers.

Now 3^{3} = 27

Can be represented by the following numbers :

 93 and 39 They hold for 2 more numbers so we have got 15 numbers till now.

Now 4^{3} = 64

This has an interesting case..

If we look at its factors, they are :

1,64

2,32

4,16

8,8

We notice that the only way of getting a two digit number which products up to 64 is 8 and 8 which is not our solution as the numbers aren't distinct.

Therefore, we find that there are 15 such numbers which qualify the conditions put up by the question.

These 15 numbers are:

10,18, 20, 24, 30, 39, 40, 42, 50, 60, 70, 80, 81, 90, 93.

Hope this was helpful.

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