Question

Three 3-digit numbers are formed using the digits 1 through 9 exactly once each. The hundreds digit of the first number is 1. The tens digit of the second number is 8. The units digit of the third number is 5. The ratio of the first number to the second number to the third number is 1:3:5, respectively. What is the sum of the three numbers?

Answer 1

1161

Explanation:

We can solve this using what we know about multiples of 3 and 5. Multiplying the first number by 3 will give the value ...

1?? × 3 = 38? . . . . where ? is an unknown digit

The two least significant digits in the first number must be ones that give 80-something when multiplied by 3. They will be between 80/3 = 26 2/3 and 90/3 = 30. The requirement that the least significant digit be odd* means the choices are 27 or 29.

When the first number is 127, its triple is 381. The 1 in the least significant digit is disallowed by the requirement that the 9 digits of these numbers be unique. (127 already has 1 as its most-significant digit.)

So, the first number is 129, making the second number be 3×129 = 387, and the third number be 5×129 = 645.

The numbers have the ratios ...

129 : 387 : 645 = 1 : 3 : 5 . . . . . . . as required.

We note that the digits 1–9 are used exactly once each, as required.

The sum of the numbers is 1161.

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* Only odd numbers will give a product that ends in 5 when they are multiplied by 5. The requirement that the third number have a ones digit of 5 means the ones digit of the first number must be odd.