What a delightful little problem ! (Partly because I could see
right away how to do it, and had the answer in a few minutes,
after a lot of impressive-looking algebra on my scratch-paper.)
Three consecutive integers are . . . x, x+1, and x+2
The smallest two are . . . x and x+1
Their product is . . . . . x(x+1)
5 times the largest one is . . . 5(x+2)
5 less than that is . . . . . . 5(x+2)-5
Now, the conditions of the problem say that <u>x (x + 1) = 5 (x+2) - 5</u>
THAT's the equation we have to solve, to find 'x' .
Eliminate parentheses: x² + x = 5x + 10 - 5
Combine like terms: x² + x = 5x + 5
Subtract 5x from each side: x² - 4x = 5
Subtract 5 from each side: <u>x² - 4x - 5 = 0</u>
You could solve that by factoring it, or use the quadratic equation.
Factored, it says that (x + 1) (x - 5) = 0
From which <em>x = -1</em>
and <em>x = +5</em>
We only want the positive results, so our three consecutive integers are
5, 6, and 7 .
To answer the question, the smallest one is <em><u>5 </u></em>.
<u>Check</u>:
5 x 6 ? = ? (7 x 5) - 5
30 ? = ? (35) - 5
30 = 30
yay !
he magician starts with the birthday boy and moves clockwise, passing out 100100100100 pieces of paper numbered 1111 through 100100100100. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer 1111 through 100100100100, and chooses the volunteer with that number.
Method2: The magician starts with the birthday boy and moves counter-clockwise, passing out 75757575 pieces of paper numbered 1111 through 75757575. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer 1111 through 75757575, and chooses the volunteer with that number.
Method 3\: The magician starts with the birthday boy and moves clockwise, passing out 30303030 pieces of paper numbered 1111 through 30303030. He cycles around the circle until all the pieces are distributed. He gives #1111 to the birthday boy, #2222 to the next kid, and so on. He then counts the number of windows in the room and chooses the volunteer with that number.
yes probabilites can be used to make fair ones
thanx
heya
Answer:
noseeee
Step-by-step explanation:
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