# What is an alternative way to calculate $36^6-26^6$?

Each user on a computer system has a password, which is six to eight characters long, where each character is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there?

P6 = $36^{6} − 26^{6}$ = 2,176,782,336 − 308,915,776 = 1,867,866,560.

Similarly, we have

P7 = $36^{7} − 26^{7}$ = 78,364,164,096 − 8,031,810,176 = 70,332,353,920

and P8 = $36^{8} − 26^{8}$ = 2,821,109,907,456 − 208,827,064,576 = 2,612,282,842,880.

Consequently,

P = P6 + P7 + P8 = 2,684,483,063,360.

My question is, instead of using the technique to find out P6 above, is there a slower way to do so without using subtracting full values? I ask this mainly for the purpose of solidifying my understanding of counting; in practice, I would prefer the technique above. thanks!!

• If you are asking for the purpose of solidifying your understanding, shouldn't you be asking for smarter, easier ways to count? Why do you care about more stupid, harder ways? – mathguy Jan 20 '17 at 3:18
• You want to simplify calculations?? – user371838 Jan 20 '17 at 3:22
• @ mathguy The smartest way doesn't necessarily allow me to look under the hook, for example, you can follow the best recipe to cook something, but it doesn't cover the details of the ingredient, etc. I just want to look at the problem from different points of view. – DSL Jan 20 '17 at 3:22
• @Rohan I want to see if there is a way to calculate the same result for P6 without using subtraction like above. Like by the product rules, instead of the sum rule. – DSL Jan 20 '17 at 3:25
• You may consider using formulas for factoring the difference of same powers such as at the following site: themathpage.com/alg/difference-two-squares-2.htm – John Wayland Bales Jan 20 '17 at 3:42

There is one way I can think of on the fly.

$36^6-26^6$

$=(36^3)^2-(26^3)^2$

$=(36^3+26^3)(36^3-26^3)$

$=(36+26)(36^2+26^2-26×36)(36-26)(36^2+26^2+36×26)$

$=62×1036×10×2908$

$=1867866560$