Grades 11-12 Video Solutions 2024-2024_11-12

Video Transcription

I write down a four-digit non-zero number, n equals pqrs.  When I place a decimal point between q and r, I find that the resulting number, pq dot rs, is the average of the two-digit numbers, pq and rs. What is the sum of the digits of n?  Let a be the first two digits combined, and b be the second two. Then, by the given relation, we know that a plus .01b is half a plus b.  We can solve and simplify to find 50a equals 49b. Then, we know that a must be 49 and b must be 50, so the sum of the digits of n is 4 plus 9 plus 5 plus 0, which is 18.

Video Summary

The problem involves finding the sum of the digits of a four-digit number \( n \), represented as \( pqrs \). When a decimal is placed between \( q \) and \( r \), \( pq.rs \) becomes the average of the two-digit numbers \( pq \) and \( rs \). Setting \( a = pq \) and \( b = rs \), the equation \( a + 0.01b = \frac{a+b}{2} \) is used, leading to \( 50a = 49b \). Solving gives \( a = 49 \) and \( b = 50 \), making the number \( 4950 \). Thus, the sum of its digits is \( 4 + 9 + 5 + 0 = 18 \).

Keywords

digit sum

four-digit number

average

equation

4950