Grades 11-12 Video Solutions 2022-2022_11-12

Video Transcription

How many positive 3-digit integers are there that are equal to 5 times the product of their digits?  So let's try to find a number n that makes this work. I'm going to write it as ABC with a line above it to indicate  that it has a hundredth digit of a, a tens digit of b, and a ones digit of c. And so now we know that it's equal to 5 times the product of their digits.  So n has to be equal to 5 ABC.  So that means that n is a multiple of 5, which means that it either ends in a 5 or a 0.  If it ends in a 0, the entire product here is 0, and that doesn't work. So that means it ends in a 5, so c has to be 5.  Okay, so now that means that n is a multiple of 25, because there are two 5s here.  That means it ends in either 7 5 or 2 5.  But now take a look at this. Since it ends in a 5, we know that it's odd,  and it's also a multiple of all of its digits. So that means none of its digits can be even. So ending in 2 5 is ruled out. So that means it must end in 7 5, like this.  But now we can just solve for a with this equation. We have 100 a plus 75 is equal to 175 a. Then we get that 75 a equals 75, so a equals 1. And so n equals 175 does the job.  And so that means there's exactly one number that does this.

Video Summary

There is exactly one positive 3-digit integer that is equal to 5 times the product of its digits, which is 175. By analyzing the structure and properties of the number, it was determined that it must end in 5, making the other digit choices limited. The process involved recognizing that the number had to be odd, a multiple of 25, and no digit could be even, ultimately leading to the solution of 175 for fulfilling the given condition.

Keywords

3-digit integer

product of digits

multiple of 25

odd number

number 175