## Question 18

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

## 1 Comment »

1. ### examsaid

(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.