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805+ (Hard)|   Algebra|         
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IMO D

Given : a^2-b^2 = 945 = (a+b)(a-b)
assuming a+b=x, a-b=y
x.y=945
Then Prime factorization of 945 = 5x7x3x3x3
No of positive divisors = (1+1)(1+1)(3+1)=16
Now x&y can be any divisors (positive or negative), so pairs will be doubled = 32
Hence for a & b, there will be 32 ordered pairs
Bunuel
If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?

A. 8
B. 16
C. 24
D. 32
E. 64


 


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If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?

\(a^2 - b^2 = (a+b)(a-b) = 945 = 3^3*5*7\)

Number of factors of 945 = (3+1)(1+1)(1+1) = 4*2*2 = 16

Let a factor be x; y = 945/x

(a+b)(a-b) = x*y =945
a+b = x; a-b =y; a = (x+y)/2; b = (x-y)/2
a-b = x; a+b = y; a = (x+y)/2; b = (y-x)/2

For each factor x, there are 2 ordered pairs.

Since there are 2 ordered pairs for each factor, the number of ordered pairs = 2*16 = 32

IMO D
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a^2-b^2 = (a-b)(a+b)
945 = 3^3*5*7
Thus Positive divisor of 945 = (3+1)(1+1)(1+1) = 16
Similarly negative divisors = 16
Thus total of 32 pairs of solution for (a,b) .
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945= a2 - b2
Prime factorize 945
= 3^3 x 5 x 7
So total factors = add 1 to power of each prime factor and multiply
So = 4 x 2 x 2= 16
We have to consider negative numbers also and - x - = +
so just double the factors
16 x 2
= 32
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prime factorization of 945 = 5 . 3 .3.3.7
=5^1 .3^3 .7^1
=2 . 4.2
=16
there are 16 (a,b) to divide 945 into 2 factors . so total number is = 16.2 =32

ans is 32
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a^2 -b^2 = 945
(a+b)(a-b)=945
Let, x=a-b & y=a+b
a=(x+y)/2. &. b=(y-x)/2 (x,y are both odd or both even to be completely divisible by 2)
xy=945=3^3*5*7 =(-x)(-y)
Total positive divisors of 945= 4*2*2=16
Total negative divisors of 945= 4*2*2=16
Total number of ordered pairs (x,y)= 16+16=32
Total ordered pair (a,b) =32

D. 32
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Answer: D) 32

a^2 - b^2 = (a+b)(a-b)
PF of 945 = 3^3 * 5 * 7
Total number of factors = (3+1)(1+1)(1+1) = 16
However, for every positive factor pair we also have the possibility of having a negative factor pair = 16(2) = 32 possibilities
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a^2 - b^2 = 945
(a+b)(a-b)=945

Let (a+b) = x , (a-b) = y

x*y=945
To see how may pairs we can split it into we need to find prime factorisation
945 = 9 x 105 = 9 x 5 x 21 = (3^3) x (5^1) x (7^1)
Total factors = (3+1)*(1+1)*(1+1) = 16

Pairs = (Factors/2) = 8

Since its finally squared we can have positive pairs, negative pairs and pos+neg pairs / neg+pos pairs
8 pairs x 4 ways = 32
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This question could be solved using factors.

as we know given; (a-b) (a+b) = 945
And the properties of factors state that when total number of factors are arranged in increasing order then the product of equidistant factors gives the number itself(you could prove it as well)

Factors of 945 are: prime factorization of 945 = 3^3 * 5 * 1
Number of factors = (3+1)*(1+1)*(1+1) = 4 * 2 * 2 = 16

So ordered pairs are asked so values of a and b could be interchanged. So there are total 8 pairs but since values could be interchanged then there would be 8 more pairs, 16 pairs possible

Also in such questions, we should check possibility of negative integers as well.

So 16 positive pairs are there and 16 negative pairs are there.

Hence answer would be 32
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From a^2-b^2 = 945, we can simplify the form to (a-b)(a+b) = 945.

Then, finding the prime factorization of 945, we've got 3^3, 5, and 7, with (3+1)(1+1)(1+1) = 16 ways to combine these numbers.

Next, we investigate (a-b)(a+b) to see what would be required to form the pair.
xy = 945
x = a-b
y = a+b

So, solving for a, x+y = 2a.

Now we know that any value that we want to combine comes from (a-b) and (a+b) pairs; they must be something that can create an even number. (2a)

There are 2 cases: when both items, (a-b) and (a+b), are odd or even.

Back to the prime factors: 3^3, 5, and 7.

We know that any odd value multiplied by an odd value gives an odd.

Which means that each item must be an odd number, which our factors are already odd.

Next, we just need to account for the negative value, so 16*2 = 32.

Choice D.
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(a+b) (a-b) = 945

lets say 1*945 = 945

a+b = 945
+ a-b = 1
= 2a = 946

a = 473, b = 472.

945 = 5*7*3*3*3

Every factor of 945 gives (a,b)

so total no. of factors is 2*2*4 = 16

Negative factors = 16. Total = 32.
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a^2 - b^2 = (a-b)(a+b) = 945
x = a-b and y = a + b, then xy = 945
also x+y = 2a or a = x+y/2
and y-x = 2b or b = y-x/2
a,b are integers
since 945 is odd, x and y are odd
945 = 3^3 * 5 * 7
no of factors = 4*2*2 = 16
each pair will have corresponding paris
(f,945/f)
so 16*2 = 32
D
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i am going with option d.
(a-b)(a+b) = 945
factors - 945 = 3^3 x 5^1 x 7^1 = 3+1, 1+1, 1+1 = 4x2x2 = 16
there are 16 positive and negative factors of x = 16+16=32, unique ordered pairs.
a incorrect - only count pairs where a and b are positive and a>b
b incorrect - only count cases where x is a positive factor, ignoring negative.
c incorrect - does not relate to factor count of 945
e incorrect - occurs if 945 had twice as many factors
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i am guessing it as 32, here is my thought

945 Prime factorized gives - 3^3 * 5^1 * 7*1
and a^2 - b^2 = (a+b)(a-b)

suppose a + b= 27, a - b = 35. then a becomes - 31 and b becomes -4
and if a+b = 35 then a-b = 27 then a becomes - 31 and b becomes 4
so two different pairs (31,4) and (31,-4)

for possible two multiple from 3^3 * 5^1 * 7^1. there are two sets of pairs
so total factors = (3+1) (1+1) (1+1) = 16
and two times = 32 so total 32 ordered pairs
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(a-b)(a+b)=945

945=3^3*5*7=16 divisor=8 positive factor pairs
all pairs are odd-- all valid
8*4=32
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Bunuel
If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?

A. 8
B. 16
C. 24
D. 32
E. 64


 


This question was provided by GMAT Club
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945 = 5 * 3 * 3 * 3 * 7

945 = (a+b)(a-b)

Number of positive divisors = 4 *2 *2 = 16

Given we have negative divisors as well. Total number of ordered pairs = 16 + 16 = 32

Option D
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It took me almost 5 minutes to do this. Can someone help with a faster method? Here is how I did it:

Given that 945 = a^2 - b^2 = (a+b)(a-b)
both a and b are integers.

Now this above info tells us that 945 can be expressed as products of 2 quantities. These 2 quantities are (a+b) and (a-b). Since and b are integers, we know that a+b and a-b will be integers as well.
Thus, essentially it is the product of two factors of 945.
Let, a+b = p and a-b = q

Therefore, 945 = p*q
Since, 945 is not a perfect square, it has even number of factors.
The total number of factors of 945 can be found and when divided by 2, we can get number of pairs of p and q.
945 = 3^3*5*7
Thus no. of factors is (3+1)(1+1)(1+1) = 4*2*2 = 16
Thus pair p and q is 16/2 = 8

I almost fell into a trap here. The above formula only gives us the number of positive factors. For 8 pairs of positive p and q, we will also have 8 pairs of negative p and q (eg: 27*35 = 945 and (-27)*(-35) = 945 as well)

Thus, total number of pairs of p and q will 16.

now we will need to find a and b.
since p=a+b and q=a-b, lets try a few pairs of factors to see if we see any patterns
Let, p = 27 and q = 35
thus, a+b =27
a-b = 35
If we solve the above equations, we get a = 36, b. = -9. So for a single pair of p and q, we get a single corresponding value of a and b

I tried a couple of pairs just to be sure. You can try that as well.

Thus, for 16 pairs of p and q, we get 16 corresponding pairs of a and b.

I almost fell for another trap here. Since we are asked for ordered pairs, we will need to multiply the above number (16) by 2 to get the final answer which is 32.

Thus, (D) 32 is the answer


Please let me know if there is a faster method to do this
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