Respuesta :
Answer:
[tex]\left\begin{array}{ccccc}&0&0&0&0 \\ \times&&&&9 \\--&--&--&--&--\\& 0 & 0 & 0 & 0 \\--&--&--&--&-- \end{array}\right[/tex]
Step-by-step explanation:
Now,
[tex]\left\begin{array}{ccccc}&M&A&T&H \\ \times&&&&9 \\--&--&--&--&--\\& H & T & A & M \\--&--&--&--&--\\ ^{a} & ^{b} & ^{c} & ^{d} \end{array}\right[/tex]
where the small letters a,b, c and d refers to carried numbers.
Converting this to equations:
- 9H=10d+M
- 9T+d=10c+A
- 9A+c=10b+T
- 9M+b=H
This equation comes from carrying remainders.
We know [tex]0 \le H \le 9[/tex] because H is a single digit, so by the fourth equation,
9M+b=H
M=0 or 1 since any larger integer value of M will make H a double digit number.
If M=1, we have:
[tex]\left\begin{array}{ccccc}&1&A&T&H \\ \times&&&&9 \\--&--&--&--&--\\& H & T & A & 1 \\--&--&--&--&--\end{array}\right[/tex]
The only multiple of 9 that will give a remainder of 1 is 9. Therefore:
In this case, H=9
However, even if A and T were to be 0, the least possible number
1009 X 9= 9081
Therefore:
M=0
And sure enough
[tex]\left\begin{array}{ccccc}&0&0&0&0 \\ \times&&&&9 \\--&--&--&--&--\\& 0 & 0 & 0 & 0 \\--&--&--&--&-- \end{array}\right[/tex]
