# binary value

how to obtain binary of following an integer.as the object is very large 189866136719308462018271159242437168532L

binary value

1

One can convert this "long" to a "Sage integer", and use the method "bits".

```
sage: a = 189866136719308462018271159242437168532L
sage: b = ZZ(a).bits()
```

How many bits do we have:

```
sage: len(b)
128
```

Show the first eight bits

```
sage: b[:8]
[0, 0, 1, 0, 1, 0, 0, 1]
```

and the last eight bits

```
sage: b[-8:]
[0, 1, 1, 1, 0, 0, 0, 1]
```

1

Alternatively, if we want the binary representation as a p h o n e number:

```
sage: a = 189866136719308462018271159242437168532L
sage: bin(a)
'0b10001110110101101110001101000111101110110011101001111001010101010010100111110111010100011010100111001000110100000000110110010100'
sage: type(bin(a))
<type 'str'>
```

This is similar to

```
sage: hex(a)
'0x8ed6e347bb3a795529f751a9c8d00d94L'
sage: oct(a)
'02166556150756635171252247672432471064006624L'
```

We can check that this matches the `bits`

solution from the answer of slelievre:

```
sage: ZZ(a).bits() == [ int(s) for s in bin(a)[:1:-1] ]
True
```

Of course.

**Note**: Repeated times i tried to post versions of this answer. (The solution was to take a shower, then i wrote the word phone in an "unexpected" way. It seems that we cannot post the string pho.. number followed by an obvious such number here.)

@dan_fulea -- regarding pho.. ..mber, yes, this is one of the strings that cause questions or answers to be rejected automatically, as it frequently signals spam. It would be nice if that could be bypassed for users with sufficient karma, but I'm not sure the Askbot engine allows that.

Please start posting anonymously - your entry will be published after you log in or create a new account.

Asked: ** 2018-04-06 19:16:46 +0200 **

Seen: **1,212 times**

Last updated: **Apr 07 '18**

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(I tried repeated times to post the answer, got internal error, that i should report...)