How do you solve for this? Your choices are: a) 5, b) 8, c) 11, d) 18.
2006-07-13 08:25:14 · 5 answers · asked by inhisgrip4eva 1 in Science & Mathematics ➔ Mathematics
Look for maximums. Well, the highest number z could be is 5. The highest number y could be is 7, and the highest number x could be is 12. If any of those are higher than the maximum, the product would be higher than 6,000. Z has the fewest possible values, so let's try there first.
Now, 3^y times 5^z must be less than half of 6,000 (since x must be at least 1), so z=5 is out. If z=4, then 5^z = 625. But since 2^x must be at least 2, and 3^y must be at least 3, the total if z=4 must be 2 * 3 * 625 = 3,750. Raising x by 1 multiplies this by 2, and we're out of range. So z=4 is out.
If z = 3, 5^z = 125. If x=1 and y=1,
2*3*125 = 750.
If we make x=5 instead, we have
16*3*125 = 6,000. I'm assuming this doesn't count, and there's no other combination of 2's and 3's you can use to get 125 to between 5,000 and 6,000.
If z=2, 5^z = 125. If x =1 and y =1,
2*3*25 = 150.
What combination of 2's and 3's could you multiply this by to get between 5,000 and 6,000? Trial and error gives you
8 * 27 * 25 = 5,400.
2^3 + 3^3 + 5^2 = 5,400.
Look at the exponents. 3 + 3 + 2 = 8.
2006-07-13 08:41:27 · answer #1 · answered by -j. 7 · 3⤊ 0⤋
Let's try 2²3²5. That's 4*9*5=180. Now we need to multiply this by a number that has zero factors of 2, 3 or 5 in order for 180 to be a prime factor AND get a result between 5000 and 6000. Let's try... (drumroll please....) 7. The next prime number.
1260. Nope. And if we multiply by 7 again, we get something too large.
Let's try ... 11. 1980.
13: 2340
skip to 29: we arrive at 5220.
Perfect. (2²3²5) and 29 are mutually prime and have a product, 5220, that lies between 5000 and 6000.
a) is the solution.
I'm really not sure if b c or d won't work, but hey, I hit it on the first try, so I'm stopping while ahead.
-edit-
I see mathgirl came up with a different solution. I think this question is flawed.
...
and now her answer is gone. Curiouser and curiouser.
2006-07-13 15:39:11 · answer #2 · answered by bequalming 5 · 0⤊ 0⤋
A) 5
2006-07-13 16:16:12 · answer #3 · answered by Brandy O 3 · 0⤊ 0⤋
I played around with the powers and tried several combinations:
(2^3)(3^2)(5^2) = 5400
Answer is B.
2006-07-13 15:42:17 · answer #4 · answered by Jeff U 4 · 0⤊ 0⤋
i think your question should read 2^x*,3^y*,5^z are the primefactors of a no between 5000and 6000.find x+y+z=8
the factore are 2^3,3^3and5^2 and the product gives me
8*27*25=5400
2006-07-13 15:43:24 · answer #5 · answered by raj 7 · 0⤊ 0⤋