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3 years ago

Simplify this algebraic expression, 5(b-4)+25=5x6-5x4

Mathematics

2 answers:

nadya68 [22]3 years ago

8 0

B=1 5(b-4)+25=5x6-5x4 5b-20+25=30-20 5b+5=10 5b=5 B=1

trapecia [35]3 years ago

4 0

5(b - 4) + 25 = (5 × 6) - (5 × 4)
5(b) - 5(4) + 25 = 30 - 20
5b - 20 + 25 = 10
5b + 5 = 10
 -5 -5
 5b = 5
 5 5
 b = 1

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All digits in a dropdown number are different, and one of its digits is the average of all its digits. It has at least two digit

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1. Start the search among 2-digit numbers. A dropdown number (DDN) with 2 digits is a number $ab$ such that

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but both of these solutions violate the requirement that the digits are distinct, so there are no 2-digit DDNs.

A 3-digit DDN $abc$ is such that

$\dfrac{a+b+c}3 = a \implies a+b+c = 3a \implies b+c = 2a$

or $a+c=2b$ if the average is $b$ , or $a+b=2c$ if the average is $c$ . The smallest possible value for $a$ is 1 since we require 3 digits. Then $b+c=2$ , and we can pick $b=0$ and $c=2$ to get the smallest DDN, 102.

2. In a 4-digit DDN $abcd$ , we have

$\dfrac{a+b+c+d}4 = a \implies a + b + c + d = 4a \implies b+c+d=3a$

or $a+c+d=3b$ or $a+b+d=3c$ or $a+b+c=3d$ .

We're free to fix $a=1$ and $b=0$ to try to get the smallest DDN. This leaves us with $c+d=3$ or $c+d=-1$ or $1+d=3c$ or $c=3d$ .

The first two cases are impossible - the only choices for $c,d$ such that $c+d=3$ are 1 and 2, and the sum of two positive integers must be positive. The smallest possible value of $c$ is 2; this leaves us with $1+d=6$ or $2=3d$ , but the latter case is impossible because 3 does not divide 2. So $d=5$ , and the smallest 4-digit DDN is 1025.

To find the largest DDN, start with the largest possible values for $a$ and $b$ . Let $a=9$ and $b=8$ . Then $c+d=19$ or $c+d=15$ or $17+d=3c$ or $17+c=3d$ . At most, we can have $c+d=13$ with 7 and 6, so the first two cases are impossible. If we maximize $c=7$ , then either $17+d=21\implies d=4$ or $24=3d\implies d=8$ (which we don't want). So the largest 3-digit DDN is 9874.