An N-bit integer holds 2N different values.
For an unsigned integer, the range of values is 0..2N-1 thus.
For a signed integer using 2s complement, the range is -2N-1..+2N-1-1.
Therefore, the largest positive number that can be stored using 8 bits is 255.
Using those 4 digits 6451 is the largest odd number you can make.
99 is the largest two digit number. If I can use the digits anyway I want, then 9 to the 9th power is larger
The greatest possible 'length' comes from the number with the greatest number of prime factors. The greatest number of factors is created by using the smallest prime number, 2, as a factor as many times as possible. Since 2^9=512 and 2^10=1024, the greatest possible 'length' of a positive integer less than 1000 is 9.
It is 998316.Actually, it is possible to make 999318 - if you turn the 6 card over so that it reads 9.
11b which is 1*2 + 1*1 = 3 would be for two bits. But a byte is 8 bits, so 2 bytes is 16 bits. The largest binary number is [2^16 - 1], which is 65535 (base ten)
674
1111 1111 1111 1111 = 2^16 = 65536
Using the digits 12279, the largest number that can be represented is 97,221.
You get the largest number if you sort the digits, from largest to smallest.
There is no largest number. You can just keep going.
0.0000001
The largest even number that can be made with those digits is 9740 .
The largest odd number that can be made with those digits is 9407 .
Arranging them in the order 9521 produces the largest number. Using operands, using the exponential function (if available) would create the largest number. [(2 ^5)^9]^1= 35184372088832
7654321
987,654,312
7542