Here is a famous old multiplication idea. Look at the two simple problems below.

```
51 21
× 3 × 6
153 126
```

The **strange** property here is that the digits that were used in the factors are the only ones that appear in the product! And even stranger is the fact that try as hard as you can, you cannot find anymore cases like this—for THREE digits, that is. But when you go to four or five digits, well, that’s a different story, as you will soon see.

In the problems below some of the digits are missing, and *** are put in their places. But they all obey this rule: * the digits used in the factors are the ONLY ones that show up in the product*, and vice versa. So for this lesson you are supposed to find the missing digits.

A) 473 B) *** C) 4128 × 8 × 9 × 3 **** 3159 ***** D) **** E) **** F) **** × 3 × 6 × 8 12843 15246 39784 G) 7*** H) **2* I) **51 × 9 × 3 × * 6*149 *1375 2*7*3 J) **14 K) **** L) **** × * × * × * *7*28 17482 51268

Now we will do the same idea, but here we will just use two-place numbers. Below are the only eight numbers that you are supposed to use to make four problems. How they should be paired off and still obey our rule of

*same digits in the product*? And what are the four correct products?

15 21 81 93 35 87 27 41

Source for idea: H. E. Dudeney, Amusements in Mathematics. Dover, 1958. pp. 15 & 156.

tt(1/23/77)