Actual Twitter Problem from 9-23-14 had additional restrictions which didn't fit in the title:

**Explain why (3^52)(7^18)(11) is the smallest positive integer with 2014 factors and which doesn't end in 5 or an even digit.**

Before the solution, a few

REFLECTIONS...

1) This is not an SAT or typical Common Core Problem. It's more challenging than that. But it does apply a fundamental principle of arithmetic which is often overlooked.

2) The solution below is more detailed than most but these are the kinds of solutions I will be emailing to you when you subscribe. See details at top of sidebar to the right.

REFLECTIONS...

1) This is not an SAT or typical Common Core Problem. It's more challenging than that. But it does apply a fundamental principle of arithmetic which is often overlooked.

2) The solution below is more detailed than most but these are the kinds of solutions I will be emailing to you when you subscribe. See details at top of sidebar to the right.

Very Very Very Detailed Solution:

There's a fundamental rule about the number of factors of any positive integer > 1. I'll demo it with 12...

Step 1. Prime factorization of 12 is (2^2)•(3^1)

Step 2. Each factor of 12 is then of the form (2^a)(3^b) where a=0,1,2 and b=0,1

Step 3. Using the multiplication principle of counting there are (3)(2)=6 possible combinations of the exponents, each one producing a unique factor of 12:

1=(2^0)(3^0) (0,0) pair

2=(2^1)(3^0) (1,0) pair

3=(2^0)(3^1) (0,1) pair etc...

4=(2^2)(3^0)

6=(2^1)(3^1)

12=(2^2)(3^1)

There's a fundamental rule about the number of factors of any positive integer > 1. I'll demo it with 12...

Step 1. Prime factorization of 12 is (2^2)•(3^1)

Step 2. Each factor of 12 is then of the form (2^a)(3^b) where a=0,1,2 and b=0,1

Step 3. Using the multiplication principle of counting there are (3)(2)=6 possible combinations of the exponents, each one producing a unique factor of 12:

1=(2^0)(3^0) (0,0) pair

2=(2^1)(3^0) (1,0) pair

3=(2^0)(3^1) (0,1) pair etc...

4=(2^2)(3^0)

6=(2^1)(3^1)

12=(2^2)(3^1)

So think of this as the

"Add 1 to the exponents and multiply" Rule!

"Add 1 to the exponents and multiply" Rule!

Back to 2014 factors...

From Wolfram Alpha (enter "factor 2014")

2014=53•19•2

To construct an integer with this many factors we reverse the previous procedure, I.e., we SUBTRACT 1:

If p1,p2,p3 are different primes then

((p1)^52)•((p2)^18)•((p3)^1) will have

53•19•2 =2014 factors!

From Wolfram Alpha (enter "factor 2014")

2014=53•19•2

To construct an integer with this many factors we reverse the previous procedure, I.e., we SUBTRACT 1:

If p1,p2,p3 are different primes then

((p1)^52)•((p2)^18)•((p3)^1) will have

53•19•2 =2014 factors!

From the conditions we want to use the 3 smallest primes excluding 2 and 5, namely 3,7,11:

(3^52)(7^18)(11^1).

QED

(Mathematician's way of saying "I'm done!")

(3^52)(7^18)(11^1).

QED

(Mathematician's way of saying "I'm done!")

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