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Alja [10]
4 years ago
5

Which of the following is equal...please help

Mathematics
2 answers:
lord [1]4 years ago
4 0

Answer:

25

Step-by-step explanation:

1/ 5^-2

The negative in the exponent in the denominator moves it to the numerator

1/ a^-b = a^b

1 /5^-2

5^2

25

Cloud [144]4 years ago
3 0

Answer:

c

Step-by-step explanation:

1/(1/25)=25

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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
Svetradugi [14.3K]

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

\dfrac{a+b}2 = a \implies a + b = 2a \implies b = a

or

\dfrac{a+b}2 = b \implies a+b = 2b \implies a=b

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

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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 <em>smallest</em> 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 <em>largest</em> 3-digit DDN is 9874.

3. I don't have an analytical solution to this, but using brute force (program) the total count is 112.

4. It is possible; consider 1249 and 1250, with digital averages

\dfrac{1+2+4+9}4=4 \text{ and } \dfrac{1+2+5+0}4=2

which happens to be the smallest pair. (Also found with brute force.)

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2 years ago
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