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16 Sep 2014, 01:52
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
10% (02:03) correct
90%
(02:07)
wrong
based on 88
sessions
History
Date
Time
Result
Not Attempted Yet
How many different sets of positive square integers, each greater than 1, add up to 75?
A. 1
B. 4
C. 7
D. 11
E. 13
A. 1
B. 4
C. 7
D. 11
E. 13
16 Sep 2014, 01:52
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Bookmarks
Official Solution:
How many different sets of positive square integers, each greater than 1, add up to 75?
A. 1
B. 4
C. 7
D. 11
E. 13
First, lay out the possible numbers you can use in the sum. The positive square integers greater than 1 but less than 75 are 4, 9, 16, 25, 36, 49, and 64.
Now, let's create a possible set and then see whether we can adjust it. It's fairly obvious that three 25's add up to 75, so our first set is {25, 25, 25}.
We might now recall the most famous example of the Pythagorean Theorem: \(9 + 16 = 25\). So we can swap out, successively, a 25 and replace it with a 9 and a 16. With sets, order does not matter, so we get three more possible sets:
{25, 25, 9, 16}
{25, 9, 16, 9, 16}
{9, 16, 9, 16, 9, 16}
This gives us 4 sets so far. However, we can now swap out 16's for four 4's. We can do so as follows.
One possible swap for the set with one 16:
{25, 25, 9, 4, 4, 4, 4}
Two possible swaps for the set with two 16's:
{25, 9, 16, 9, 4, 4, 4, 4}
{25, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
And three possible swaps for the set with three 16's:
{9, 16, 9, 16, 9, 4, 4, 4, 4}
{9, 16, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
Before going to the larger squares, we should glance over our list and see whether we can do any swaps within the sets we've already created, using only squares equal to 25 or less. The only swap we can do is in the last set: we can swap out nine 4's and replace them with four 9's:
{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4} = three 9's and twelve 4's
becomes
{9, 9, 9, 9, 9, 9, 9, 4, 4, 4} = seven 9's and three 4's
We are now at a total of 11 sets, having exhausted the possibilities that only involve the squares equal to 25 or less. Are there any sets that involve larger squares?
We can quickly check:
64 can't be in the set, because the leftover (11) can't be formed from the sum of 9's and/or 4's.
49 CAN be in the set. The leftover (26) can be formed by the sum of two 9's, and two 4's, so we get
{49, 9, 9, 4, 4}
36 CAN be in the set. The leftover (39) can be written as the sum of three 9's and three 4's, so we get
{36, 9, 9, 9, 4, 4, 4}
Thus, the total number of different sets is 13.
Answer: E
How many different sets of positive square integers, each greater than 1, add up to 75?
A. 1
B. 4
C. 7
D. 11
E. 13
First, lay out the possible numbers you can use in the sum. The positive square integers greater than 1 but less than 75 are 4, 9, 16, 25, 36, 49, and 64.
Now, let's create a possible set and then see whether we can adjust it. It's fairly obvious that three 25's add up to 75, so our first set is {25, 25, 25}.
We might now recall the most famous example of the Pythagorean Theorem: \(9 + 16 = 25\). So we can swap out, successively, a 25 and replace it with a 9 and a 16. With sets, order does not matter, so we get three more possible sets:
{25, 25, 9, 16}
{25, 9, 16, 9, 16}
{9, 16, 9, 16, 9, 16}
This gives us 4 sets so far. However, we can now swap out 16's for four 4's. We can do so as follows.
One possible swap for the set with one 16:
{25, 25, 9, 4, 4, 4, 4}
Two possible swaps for the set with two 16's:
{25, 9, 16, 9, 4, 4, 4, 4}
{25, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
And three possible swaps for the set with three 16's:
{9, 16, 9, 16, 9, 4, 4, 4, 4}
{9, 16, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4}
Before going to the larger squares, we should glance over our list and see whether we can do any swaps within the sets we've already created, using only squares equal to 25 or less. The only swap we can do is in the last set: we can swap out nine 4's and replace them with four 9's:
{9, 4, 4, 4, 4, 9, 4, 4, 4, 4, 9, 4, 4, 4, 4} = three 9's and twelve 4's
becomes
{9, 9, 9, 9, 9, 9, 9, 4, 4, 4} = seven 9's and three 4's
We are now at a total of 11 sets, having exhausted the possibilities that only involve the squares equal to 25 or less. Are there any sets that involve larger squares?
We can quickly check:
64 can't be in the set, because the leftover (11) can't be formed from the sum of 9's and/or 4's.
49 CAN be in the set. The leftover (26) can be formed by the sum of two 9's, and two 4's, so we get
{49, 9, 9, 4, 4}
36 CAN be in the set. The leftover (39) can be written as the sum of three 9's and three 4's, so we get
{36, 9, 9, 9, 4, 4, 4}
Thus, the total number of different sets is 13.
Answer: E
23 Mar 2015, 01:53
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this should be 700+ difficulty as it is taking so much time....
Any shortcut for this?
Any shortcut for this?
