A number has 3 prime factors. The number is greater than 25 but less than 35. If no factor is repeated, what is the number and what are the factors?
Rod has 8 coins in his pocket. He has no half dollars. The value of the coins is $1.17. What are the coins?
Thank you
2007-12-16 14:40:03 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x is a number we are looking for.
25
It has 3 prime factors.
lets list a few prime numbers:
2,3,5,7,11,13,...
2x3x5 = 30 would be your number (3 prime factors, non of them repeating)
Sorry, but I have no idea what coins are there when it comes to dollars. Sorry,
Kempos
2007-12-16 14:44:38 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The three factors:
2 * 3 * 5 = 30
Coins in Rod's pocket:
Answer: 2 pennies, 1 nickel, 1 dime and 4 quarters—8 is the number of coins and the value is $1.17.
2007-12-16 22:54:39 · answer #2 · answered by Jun Agruda 7 · 3⤊ 0⤋
1. Prime factors are 2, 3, and 5, multiplied together to equal 30. They're "prime" because they're only divisible by one and themselves.
2. 4 quarters, a dime, a nickel, and 2 pennies
2007-12-16 22:45:39 · answer #3 · answered by savethepenguins123 2 · 0⤊ 0⤋
The answer to the first is 30. The factors of 30 are 2,3,5.
The second is four quarters, a dime, a nickel, and two pennies.
2007-12-16 22:47:48 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Dont know the answer to the first but second is 4 quarters, a dime a nickel and two pennies...
2007-12-16 22:44:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
30-----3,2,5
$1.17= 4 quarters 1 dime 1 nickel 2 pennies
2007-12-16 22:46:34 · answer #6 · answered by Bennie Y 2 · 0⤊ 0⤋
i do not know the answer to your first question but , i am guessing for you second question about the $ 1 . 17 .4 quaters a dime a nickel and two pennies.
2007-12-16 22:52:54 · answer #7 · answered by rcsmith52107 2 · 0⤊ 0⤋
Prime factors eligible: 1,2,3,5,7...
Eliminate 1 & 7, leaves 2,3,5.
Product is 30.
2 must be pennies,
4 could be quarters,
1 nickle, & 1 dime.
2007-12-16 22:49:24 · answer #8 · answered by Robert S 7 · 0⤊ 0⤋