I dont understand this one.... is the answer 4?prachich1987 wrote:How many different sets of positive square integers, each greater than 1, add up to 75?
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The numbers must be less than 75 but greater than 1.prachich1987 wrote:How many different sets of positive square integers, each greater than 1, add up to 75?
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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: . 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.
The correct answer is E.
Isn't it a lengthy & time consuming question?Anurag@Gurome wrote:The numbers must be less than 75 but greater than 1.prachich1987 wrote:How many different sets of positive square integers, each greater than 1, add up to 75?
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There are only seven such square numbers: 4, 9, 16, 25, 36, 49, and 64.
A fairly obvious set is (25, 25, 25). ONE
Now 25 can be written as the sum of two squares as (16 + 9). Hence we have the opportunity to break each 25 into two more squares. Which results in another three possible set: (9, 16, 25, 25), (9, 9, 16, 16, 25) and (9, 9, 9, 16, 16, 16). THREE
Now 16 can be written as the sum of four squares as (4 + 4 + 4 + 4). Hence we have the opportunity to break each 16 into four more squares. Which results in another six possible set: (9, 16, 25, 25) gives 1, (9, 9, 16, 16, 25) gives 2 and (9, 9, 9, 16, 16, 16) gives 3. SIX
Now whenever I have more than nine 4's in a set, we can replace them with four 9's. We have twelve 4's when all the 16's are broken into 4's. Thus nine 4's can be replaced with four 9's resulting a new set. ONE
Again nine 4's or four nine's can be replaced by a square number 36 resulting a new set. ONE
Note that when one 16 of the set (9, 9, 16, 16, 25) is broken into four 4's the set will be like (4, 4, 4, 4, 9, 9, 16, 25). Now we can combine one 16, two 4's and one 25 to get 49 resulting a new set. ONE
Thus total number of possible sets = (1 + 3 + 6 + 1 + 1 + 1) = 13
The correct answer is E.
Anurag@Gurome wrote:The numbers must be less than 75 but greater than 1.prachich1987 wrote:How many different sets of positive square integers, each greater than 1, add up to 75?
1
4
7
11
13
There are only seven such square numbers: 4, 9, 16, 25, 36, 49, and 64.
A fairly obvious set is (25, 25, 25). ONE
Now 25 can be written as the sum of two squares as (16 + 9). Hence we have the opportunity to break each 25 into two more squares. Which results in another three possible set: (9, 16, 25, 25), (9, 9, 16, 16, 25) and (9, 9, 9, 16, 16, 16). THREE
Now 16 can be written as the sum of four squares as (4 + 4 + 4 + 4). Hence we have the opportunity to break each 16 into four more squares. Which results in another six possible set: (9, 16, 25, 25) gives 1, (9, 9, 16, 16, 25) gives 2 and (9, 9, 9, 16, 16, 16) gives 3. SIX
Now whenever I have more than nine 4's in a set, we can replace them with four 9's. We have twelve 4's when all the 16's are broken into 4's. Thus nine 4's can be replaced with four 9's resulting a new set. ONE
Again nine 4's or four nine's can be replaced by a square number 36 resulting a new set. ONE
Note that when one 16 of the set (9, 9, 16, 16, 25) is broken into four 4's the set will be like (4, 4, 4, 4, 9, 9, 16, 25). Now we can combine one 16, two 4's and one 25 to get 49 resulting a new set. ONE
Thus total number of possible sets = (1 + 3 + 6 + 1 + 1 + 1) = 13
The correct answer is E.
