Question

tony is using the numbers 12 7 4 1 to make a new number
he can use brackets, +, -, ÷, × as much as often as he wishes
he cannot use any number more than once
he cannot use powers
he cannot put numbers together e.g, he can't use 147
what is the biggest number he can make show how he makes it

Answer 1

Given:

Tony is using the numbers 12 7 4 1 to make a new number

he can use brackets, +, -, ÷, × as much as often as he wishes

he cannot use any number more than once

he cannot use powers

he cannot put numbers together e.g, he can't use 147

To find:

What is the biggest number he can make show how he makes it

Solution:

From given, we have,

Tony is using the numbers 12 7 4 1 to make a new number

(multiplication of digits give a larger number compared to addition, subtraction and division)

let us consider,

= (12 × 7) × (4 + 1)

so, we have,

12 × 7 = 84

4 + 1 = 5

Therefore, we get,

84 × 5 = 420

∴ The biggest number Tony can make is 420

Answer 2

Answer:

If you can use parentheses, ^, +, -, *, and /, then 4^(7^(12+1)) is a 58,332,996,766 digit number.

If concatenation is allowed, then 4^(7^121) is a number with about 108 googol digits. To put it another way, 4^(7^121) is bigger than a googolplex.[1]

Even if concatenation isn’t allowed, but you can use any number of square root symbols, then there’s a way to make an arbitrarily large number.

Consider the following sequence

a1=12∗7/( $\sqrt{4}$ − 1 ) = 84  

a2=12∗7/($\sqrt{ \sqrt{4} }$1)≈202.8  

a3=12∗7/($\sqrt{} \sqrt{} \sqrt{4}$−1)≈444  

a4=12∗7/($\sqrt{} \sqrt{} \sqrt{} \sqrt{4}$−1)≈928.1  

etc.  

With each step, the number more than doubles, so then  $a_{googolplex}$  — a number formed using 12, 1, 4, and 7… and a googolplex square root symbols — is a number with more than three tenths of a googolplex digits.