A number is called flippy if its digits alternate between two distinct digits. For example, 2020 and 37373 are flippy, but 3883 and 123123 are not. How many five-digit flippy numbers are divisible by 15? A 3
B 4
C 5
D 6
E 8

B: 4

Step-by-step explanation:

A number is divisible by 15 if the number ends on a 0 or a 5, and the sum of its digits is divisible by 3.

We want to find a 5 digit flippy number, this can be: modeled with:

N = ababa

Where a and b are single digit numbers.

If we impose that N must end with a zero, then we will have:

N  = 0b0b0 = b0b0

This is a 4-digit number, so we can discard all the options that end with a zero.

Then the only option that we have are the numbers like:

B = 5b5b5

This number will be only divisible by 15 if:

5 + b + 5 + b + 5 = K is divisible by 3.

Then let's try find the possible values of b.

K = 3*5 + 2*b

K has two terms, the left term is already divisible by 3, then K will be divisible by 3 only if the other term is also divisible by 3.

Then we want 2*b to be divisible by 3.

And 2 is a prime number, then b must be divisible by 3, and we know that b is a number between {0,1 , 2, 3, 4, 5, 6, 7, 8 ,9}

The options are:

2*0 = 0 is divisible by 3.

2*3  = 6 is divisible by 3.

2*6 = 12 is divisible by 3

2*9 = 12 is divisible by 3.

Then we have four values of b such that:

N = 5b5b5 is divisible by 15.

Then the correct option is: B: 4

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

i think the correct option is the second option.