How is 100 the smallest 3 digit number in base seven?

Question

Question 1a: What is the smallest 3 digit number in base seven?

 

Apparently the answer is 100 but I don't know how you go about getting that from the question. All I can think of is that it's because 10^2=100. Help please?

 

Question 1b: Add the two bases together. 34 base5 + 23 base5

I tried solving this and got 112 base5 as an answer, but I'm very unsure about it. If my answer is wrong could someone please explain what I did wrong or how to properly solve this?

 

Thank you.

 

Answer 1

It works just like our (base 10) number system, except it is not powers to 10.

It's powers of the base that is specified.

CONCEPTUALLY, here is what you need to understand and remember....

Base 10, the digits are 0,1,2,3,4,5,6,7,8,9
Base 7, the digits are 0,1,2,3,4,5,6
Base 5, the digits are 0,1,2,3,4
Base 2 (computer binary) the digits are 0,1
Base 8 (octal) the digits, are 0,1,2,3,4,5,6,7
Base 16 ( hexidecimal) the digits are 0,1,2,3,4,5,6,7,8,9,A=10,B=11,C=12,D=13,E=15 <--- yes, 16 digits!

Base 4, the digits are 0,1,2,3
See a pattern?????
What are the digits in base 9?
What are the digits in base 3?

The expanded form is how we convert from one base to another.

Our base 10, for example, 123 = 100 + 20 + 3 = 1*10^2 + 2*10^1 + 3*10^0

                                     1234 = 1000 + 200 + 30 + 4 = 1*10^3+2*10^2 +3*10^1 + 4*10^0

Notice the one's place always has exponent zero.
The second place always has exponent 1.
etc etc

So for example, let's convert 123 to base 8 (octal)
8^0 = 1
8^1 = 8
8^2 = 64
8^3 = 512 <--- too big, can stop here

123 divided by 64 = 1 with a remainder of 59
59 divided by 8 = 7 with a remainder of 3
3 divided by 1 = 3 remainder of 0

So 123 in octal is 173

CHECK: 1*8^2 + 7*8^1 + 3 * 8^0 = 1*64 + 7*8 + 3*1 = 64 + 56 + 3 = 123 <-- order of operations PEMDAS
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Convert 50 to binary (base 2)

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64 <--- too big, stop here for calculuation purposes;
However, binary numbers are usually 8 bits long. So the
first two bits will be zero because we don't need them.

(Just like in base 10,  123 = 00123; you can put as

 many leading zeros as you want)

 

So, 2 leading zeros. Then....

 

50 divided by 32 = 1 remainder 18
18 divided by 16 = 1 remainder 2
2 divided by 8 = 0 remainder 2
2 divided by 4 = 0 remainder 2
2 divided by 2 = 1 remainder 0
0 divided by 1 = 0 remainder 0

50 in binary is 00110010 <-- 8 bits

CHECK: Only the 1 bits contribute, so
2^5 + 2^4 + 2^1 = 32 + 16 + 2 = 50

 

So the answer to your questions are:

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Part 1:

Base 7 means the digits are 0,1,2,3,4,5,6

Then the largest 2-digit number in base 7 is 66, which in base 10 is:

6*7^1 + 6*7^0 =
6 * 7 + 6*1 =
42 + 6 = 48.

100 in base 10 is:
1 * 7^2 + 0*7^1 + 0*7^0 =
1 * 49 + 0 + 0 = 49.

 

So yes, in base 7, 100 is the smallest 3 digit number


part 2:

base 5 means the digits are 0,1,2,3,4.
so any sum larger than 4 rolls over and forces
a carry.

so in base 5, 4 + 3 = 2 ( Think of it in base 10 as 4 + 3 = 7,
and then 7 divided by 5 is 1 with a remainder
of 2. THe REMAINDER is what stays behind. The
one gets carried over)

Next, we have 3 + 2 plus the carry. This total in base 5 is 1.
(Again, in base 10, 3 + 2 + 1 = 6, 6 divided by 5 is 1 with
a remainder of 1. So the remainder of 1 stays behind, and the final one carries over)

The final answer in base 5 is 34 + 23 = 112

Optionally you can do it like this:

First add the numbers as you would in base 10:

34 + 23 = 57.

Now 7 is not a valid digit in base 5. Remember, 7 divided by 5 is 1 '

with a remainder of 2.The correct digit is 2, and 1 carries. The answer,

including the carry, becomes 62. The 6, once again, is not a valid

digit in base 5. 6 divided by 5 is 1 with a remainder of 1. The correct

digit is 1 and the final one carries. The final answer is 112.

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Please ask more questions if you do not understand.

Answer 2

1a.

The integer smaller then 100 is 66. The number larger is 101. So 100 must be the smallest three digit number.

 

1.b

3₅ + 4₅ = 12₅     

30₅ + 20₅ =100₅

34₅ + 24₅ = 112₅

 

In base 5 the count is

0, 1, 2, 3, 4, 

10, 11, 12, 13, 14,

20, 

So for any sum 5 or greater you shall "carry" to the next place.