This is the following calculation done with no Shin method?

131-45=75

1) 6 Shin method

2) 7 Shin method

3) 8 Shin method

4) 9 Shin method

131-45=75

1) 6 Shin method

2) 7 Shin method

3) 8 Shin method

4) 9 Shin method

%d bloggers like this:

## exam said

Answer > 4

(Explanation 1) this 2000 software development engineer test morning question 1

It is remodelling problem. Originally problem was the arithmetic expression, “131-45=53”.

By the way correct answer of originally problem becomes 2. 7 Shin methods.

0s131 = 1 * 7^2 + 3 * 7 + 1 = 71

0s45 = 4 * 7 + 5 = 33

0s53 = 5 * 7 + 3 = 38

71 – 33 = 38

Is. If just the lower 1 columns you see, because it becomes 1-5=3, the 1=8 and the 0=7, in every multiple of 7

It is found that it is the carry of the beam, as for this proof of the thing which is the quantities of 7 Shin

It becomes.

It is 131-45=75, but when you think in the same way at the lower 1 columns, being 1-5=5, the 1=10 and the 0=9

So it understands that they are the quantities of 9 Shin. In addition when both sides are corrected in decimal

0n131 = 1 * 9^2 + 3 * 9 + 1 = 109

0n45 = 4 * 9 + 5 = 41

0n75 = 7 * 9 + 5 = 68

109 – 41 = 68

By the way, it is almost not used, but as for the quantities of 6 Shin as for sextal and the quantities of 7 Shin septimal,

As for the quantities of 9 Shin you call nonal.

If we assume that there is a file which is expressed at the quantity of respective Shin the senary file,

It is called the septenary file and the nonary file, probably will be, binary [hua]

Like [iru].

Furthermore 8 bits were 1 octets (octet), but 6,7,9 you call numerical value

When it makes 1 summaries, it becomes the unit, sextet, septet and nonet probably will be.