Could not understand the solution

[RamanK]

Find the smallest natural number of the form 2^a3^b7^c, such that the half of the number is the cube of an integer, one third of the number is the seventh power of an integer and one seventh of the number is square of an integer.
[Hint: 2^a 3^b 7^c/2 is a cube only if a – 1, b, c is all divisible by 3].


Given 2^a3^b7c/2 = n³ ═> 2^{a - 1} 3^b7^c = n³
2^a3^b7c/3 = m³ ═> 2^a3^{b - 1}7^c = m⁷
2^a3^b7^c/7 = l² ═> 2^a3^b7^{c - 1} = l²

a = 28 b = 36 and c = 21

Can someone please explain how 2^{a - 1} 3^b7^c will become n³?

 

[Deleted member 4993]

Find the smallest natural number of the form 2^a3^b7^c, such that the half of the number is the cube of an integer, one third of the number is the seventh power of an integer and one seventh of the number is square of an integer.
[Hint: 2^a 3^b 7^c/2 is a cube only if a – 1, b, c is all divisible by 3].


Given 2^a3^b7c/2 = n³ ═> 2^{a - 1} 3^b7^c = n³
2^a3^b7c/3 = m³ ═> 2^a3^{b - 1}7^c = m⁷
2^a3^b7^c/7 = l² ═> 2^a3^b7^{c - 1} = l²

a = 28 b = 36 and c = 21

Can someone please explain how 2^{a - 1} 3^b7^c will become n³?

Given 2^a3^b7c/2 = n³ ═> 2^{a - 1} 3^b7^c = n³
2^a3^b7c/3 = m³ ═> 2^a3^{b - 1}7^c = m⁷
2^a3^b7^c/7 = l² ═> 2^a3^b7^{c - 1} = l²

Please explain your work - including definitions of m and l.

 

[RamanK]

Given 2^a3^b7c/2 = n³ ═> 2^{a - 1} 3^b7^c = n³
2^a3^b7c/3 = m³ ═> 2^a3^{b - 1}7^c = m⁷
2^a3^b7^c/7 = l² ═> 2^a3^b7^{c - 1} = l²

Please explain your work - including definitions of m and l.

I have not solved it. I copy pasted the text book solution. If someone explain it would be helpful

 

[Deleted member 4993]

I have not solved it. I copy pasted the text book solution. If someone explain it would be helpful

Can you please give us the name/author/publisher of the textbook - so that we can look it up and explain it to you?

In the mean time,

Have you worked through the example problems of the book?

 

[Otis]

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...}

This infinite set of numbers goes by multiple names.

It's the set of counting numbers.

It's the set of Natural numbers.

It's the set of positive Integers (unless you're an electrical engineer).

The given expression (2^a*3^b*7^c) is a factored form.

Evaluating it, using Denis' values for a,b,c, yields the Natural number 457,419,312.

 

[Deleted member 4993]

Find the smallest natural number of the form 2^a3^b7^c, such that the half of the number is the cube of an integer, one third of the number is the seventh power of an integer and one seventh of the number is square of an integer.
[Hint: 2^a 3^b 7^c/2 is a cube only if a – 1, b, c is all divisible by 3].

We need the '2' for the first condition.