While raising 76 to any power the resulting value always ends with 76. This property is as same as the number 5 and 25.

• ⁷⁶² =>5776
• 76³=>438976
• 76⁴=>33362176
• 76⁵=>253552537
• 76⁶=>192699928576
• 76⁷=>14645194571776
• 76^{8 =>1113034787454976}

and goes on..

Now we are going to find the hundredth digit and thousandth digit in the result of 76ⁿ

• 76¹=> 76

• 76² =>5776 here the thousandth digit and hundredth digit is "57" By the same way the thousandth digit and hundreath digit for

• 76³=>438976 here the thousandth digit and hundredth digit is "89"

• 76⁴=>33362176 here the thousandth digit and hundredth digit is "21"

• 76⁵=>2535525376 here the thousandth digit and hundredth digit is "53"....

I think you got the pattern in which the 4th and 3rd digit getting increased.

In 76² its 57 and in the next power its 89 and in the next 21 and in the next its 53 and continues..

which is in the pattern of 57+32=>89, 89+32=>121 ,121+32=>153,153+32 =>185......

if 89+32 it gives 121 in this case consider 21 alone. These rules also applied for all the powers of 76.

so we can easily find the last four digit of the ---> 76ⁿ

provided if we know last four digit of the previous power i.e., for

---> 76^(n-1) .

So we will derive to get the last four digits without knowing the previous result.

Make sure that all these derivations applicable only for n>=2

1. if n=2=> 57(we can split it as 25+32)

2. n=3=> 57+32=>89=>(25+32+32)=>25+2*(32)=> 25+(n-1)*32

3. n=4=> 89+32=>121=>(25+32+32+32)=>25+3*(32)=> 25+(n-1)*32

4. n=5 => 21+32=>53 =>(25+32+32+32+32)=>25+4*(32)=> 25+(n-1)*32

5. n=6 => 53+32=>85 =>(25+32+32+32+32+32)=>25+5*(32)=> 25+(n-1)*32

6. n=7 => 85+32=>117 =>(25+32+32+32+32+32+32)=>25+6*(32)=> 25+(n-1)*32

and so on for all power values of n

just we found the 3rd and 4th digit. And we know the last two digits always 76

Derivation to get the general formula:

finding last four digits for 76ⁿ

=>25+(n-1)*32 by this we will get 3rd and 4th number,but may be with many digits.

To get the last two digit we make have a modulo

=>[25+(n-1)*32]%100

example:

76⁴=33362176

n=2 --> [25+(4-1)32]%100

[25+ 96]%100-> [121]%100 -> 21

to make general for last four digits multiply the above result by 100 and add 76 with it 21*100+76=2176

which is the last four digit of 76⁴

Finally the general formula for last four digit

76ⁿ=> XXXXXXXXXXXXXXXXXXXXXabcd

abcd=>[([25+32*(n-1)]%100)*100 ]+76