Question

Replace the letters with digits (the same letters represent the same digits):
MATH*9=HTAM

Answer 1

`\left\begin{array}{ccccc}&0&0&0&0 \\ *&&&&9 \\--&--&--&--&--\\& 0 & 0 & 0 & 0 \\--&--&--&--&-- \end{array}\right`

Explanation:

Now,

`\left\begin{array}{ccccc}&M&A&T&H \\ *&&&&9 \\--&--&--&--&--\\& H & T & A & M \\--&--&--&--&--\\ ^(a) & ^(b) & ^(c) & ^(d) \end{array}\right`

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
`0 \le H \le 9` 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:

`\left\begin{array}{ccccc}&1&A&T&H \\ *&&&&9 \\--&--&--&--&--\\& H & T & A & 1 \\--&--&--&--&--\end{array}\right`

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

`\left\begin{array}{ccccc}&0&0&0&0 \\ *&&&&9 \\--&--&--&--&--\\& 0 & 0 & 0 & 0 \\--&--&--&--&-- \end{array}\right`