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, 2019

Number theory Difficulty 3.8 AMC 10/12 Find the answer Canada

Rich chooses a 4-digit positive integer. He erases one of the digits of this integer. The remaining digits, in their original order, form a 3-digit positive integer. When Rich adds this 3-digit integer to the original 4-digit integer, the result is 6031. What is the sum of the digits of the original 4-digit integer?

Pick one

Solution

The single-digit divisors of 36 are: $1,2,3,4,6$ , and 9.

The groups of 3 of these digits whose product is 36 are: $1,4,9$ ; $1,6,6$ ; $2,2,9$ ; $2,3,6$ , and $3,3,4$ .

Next, we count the number of ways to arrange each of these 5 groups of digits.

The digits $1,4,9$ can be arranged to form: $149, 194, 419,491, 914, 941$ .

The digits $1,6,6$ can be arranged to form: $166, 616, 661$ .

The digits $2,2,9$ can be arranged to form: $229,292,922$ .

The digits $2,3,6$ can be arranged to form: $236,263,326,362,623,632$ .

The digits $3,3,4$ can be arranged to form: $334,343,433$ .

The number of 3-digit positive integers having digits whose product is 36 is $6+3+3+6+3=21$ .

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.