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To find the number of positive integers that satisfy both conditions, we can consider the possible combinations of digits (1, 3, 5) that add up to 6.

Let's break down the possible cases:

Three 1's: 1+1+1+1+1+1=6

One 3 and three 1's: 3+1+1+1=6 

One 5 and one 1: 5+1=6

So, there are three possible combinations that satisfy both conditions. Therefore, there are three positive integers: 111111, 3111,1311,1131,1113.15 and 51.

(a+2)(3a^2-2a-12)(a^2-2a+1)

=(3a^3-2a^2-12a+6a^2-4a-24)(a^2-2a+1)=(3a^3+4a^2-16a-24)(a^2-2a+1)=3a^5+4a^4-16a^3-24a^2-6a^4-8a^3+32a^2+48a+3a^3+4a^2-16a-24

=3a^5-2a^4-21a^3+12a^2+32a-24

The quadratic roots are 1 and -1.then, the factor of the quations are x-1 and x+1.

SO, our answer is (x-1)(x+1)=x^2-1

Laverne starts counting out loud by ${}5$'s.  She starts with $2$.  As Laverne counts, Shirley sums the numbers Laverne says.  When the sum finally exceeds $20,$ Shirley runs screaming from the room.  What number does Laverne say that sends Shirley screaming and running?  

Laverne's         Shirley's  

Counting            Sums  

        2                    2  

        7                    9  

      12                  21      I don't get it.  The running and screaming part, I mean. 

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BY triangles iniquality 

TU+UV>TVTHEN,6x-11+3x-1>2x+3 then calculating this we got x>15/7

UV+TV>TUthen,3x-1+2x+3>6x-11,then claculating this we got 13>x.

TV+TU>UV then,2x+3+6x-11>3x-1then calculating this we got x>7/5.

ANSWER IS 13>X>15/7.(THESE ARE THE POSSIBLE VALUES OF X.)

WE GOT X WHEN X MULTIPLY BY CONSTANT OR CONSTANT MULTIPLY BY X TERMS . 

HERE ,WHEN X IS MULTIPLY BY 14 THEN WE GOT 14X.

ALSO -23X IS MULTIPLY BY 1 WE GOT -23X.

-23X+14X=-7X.

SO , OUR ANSWER IS -7.

8 squared is 64 so every integer whose absolute value is less than 8 satisfies the problem. 

These are –7, –6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, 6, and 7  

I count 15 of them.  

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