1720.11 – TTT


Two two-digit numbers are being multiplied together:

YE×METTT \begin{array}{rccc} & & Y & E \\ \times & & M & E \\ \hline & T & T & T \end{array}

If each letter represents a unique digit, what is the value of E+M+T+YE + M + T + Y?

  1. 19
  2. 20
  3. 21
  4. 22
  5. 24

Solution

A good way to start is by considering EE because both YEYE and MEME end with it. Whatever the value of EE, its square determines the last digit of the product, TTTTTT, and hence the whole product.

TABLE 1 below lists all of possible values for EE together with the value of TT implied. We can rule out the values that make T=ET = E because the letters stand for unique digits. Thus we eliminate E=1,6, or 0E = 1, 6, \text{ or } 0.

TABLE 2 lists the remaining possibilities for EE, together with the whole resulting number TTTTTT and its prime factorization. From the table we see clearly that, when writing TTTTTT as a product of two two-digit numbers, one of those numbers must be 3737.

Because one of our two numbers must be 3737, we know that the other must end in 7 as well. Examining the table, we find that only E=7E = 7 works. We now know that:

YE×ME=TTT,27×37=999. \begin{aligned} YE\times ME &=& TTT, \\ 27\times 37&=&999. \end{aligned}

We were really asked for the sum E+M+T+Y=7+3+9+2=21.E+M+T+Y=7+3+9+2=21. That turns out to be 21. The answer is c.

1720_11_solution_3c714b6374.png