i, I wonder that is there any way to disassemble this operation ((n+2^(n2))^2 mod 1000000007) without counting 2^n2. Please, help me.

i, I wonder that is there any way to disassemble this operation ((n+2^(n2))^2 mod 1000000007) without counting 2^n2. Please, help me.
The modulo function, mod function (programming) is often mistaken for the word modify. All it simply does is continue to subtract/add a particular number to the current one until the number becomes between 0 and the determined value. In programming syntax, we could do. (GML) a=number b=boundary_number for ab a+=b until a<b and a>0 However, it must be noted that this shall only operate for a positive number. For this equation to work, there must be a loop sequence so that the equation can continue to canceldown the value until it reaches it's target.
"mod_pow" is a function (using fast modular exponentiation) which returns the result of exponentiation modulo 1000000007  mod_pow (the first parameter, second parameter, modulo). The first parameter is the base, and the second parameter is the exponent.
For:
y < n + mod_pow (2, n2, 1000000007).
result < mod_pow (y, 2, 1000000007).
I get an incorrect result, so I had misunderstood something. Very please to explain.
I'm not a hundred percent sure I understand your question, but in case you want to try and calculate this by hand, no, you don't have to calculate 2^(n2).
First of all, (
(n+2^(n2))^2 % 1000000007 = (n+2^(n2)) * (n+2^(n2)) % 1000000007
= ((n+2^(n2)) % 1000000007) * ((n+2^(n2)) % 1000000007) % 1000000007
= (n % 1000000007 + 2^(n2) % 1000000007)) * (n % 1000000007 + 2^(n2) % 1000000007)) % 1000000007
Now, what you have to do is calculate ((n % 1000000007 + 2^(n2) % 1000000007), square that and do the modulooperation once.
The hardest part will be calculating s = 2^(n2) % 1000000007
s = 2^(n2) % 1000000007
= ((2^30 % 1000000007) * 2^(n230)) % 1000000007
= (73741817 * 2^(n230)) % 1000000007
= (73741817 * 73741817 * 2^(n23030)) % 1000000007
etc.
HTH,
Johannes
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