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For how many unique pairs of nonnegative integers {a, b} is the equati

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For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 01 May 2015, 02:21
3
27
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A
B
C
D
E

Difficulty:

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Question Stats:

30% (02:47) correct 70% (02:45) wrong based on 253 sessions

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Re: For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 04 May 2015, 03:07
Bunuel wrote:
For how many unique pairs of nonnegative integers {a, b} is the equation a^2 - b^2 = 225 true?

A) 1
B) 3
C) 5
D) 7
E) 9


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

If a^2-b^2=225, then (a+b)(a-b)=225. The problem specifies that a and b are nonnegative, so a + b must represent the larger of the two factors and a – b must represent the smaller of the two factors. This is true in all cases but one.

What is that one case? Quick, what’s the square root of 225? If you don’t know that it’s 15, add this to your to-memorize list. The two factors could both equal 15; in this case, a = 15 and b = 0.

What are the other possible factors of 225? Break the 15s down to primes: 3 ´ 5 ´ 3 ´ 5, or 3^2´ 5^2. If a number equals 3^2´ 5^2, then the factors could be made of any combination of 3^0, 3^1, 3^2 and 5^0, 5^1, and 5^2. Because there are three options for each of two base prime numbers, there are 3 ´ 3 = 9 distinct factors for this number.

Find the lowest numbers first and then do the math to find the factor pair:
3^0*5^0=1. This must be paired with 225 (because 1 × 225 = 225).
3^1*5^0=3. This must be paired with 225 / 3 = 75.
3^0*5^1=5. This must be paired with 225 / 5 = 45.
3^2*5^0=9. This must be paired with 225 / 9 = 25.
3^1*5^1=15. This must be paired with 225 / 15 = 15.

That’s a total of 9 factors, so you know you’re done. How can you use that info to find the unique pairs of a and b?

Recall that (a+b)(a-b)=225. Take the first pair:
Attachment:
Screen-shot-2013-08-19-at-9.35.21-AM.png
Screen-shot-2013-08-19-at-9.35.21-AM.png [ 9.01 KiB | Viewed 5052 times ]

Given that a+b=225 and that a-b=1, how can you solve for a and b?

Subtract equations:

a+b=225
-(a-b)=1
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2b=(225-1)
b=(225-1)/2

In other words, the value of b is equal to (Factor 1 – Factor 2) / 2. By the same token, the value of a is equal to (Factor 1 + Factor 2) / 2. (If you’re not convinced about that, prove it using the same kind of math shown for b above.)

So, for factors 225 and 1, a={(225+1)/2}=113 and b={(225-1)/2}=112.

Test the remaining factor pairs:
Attachment:
Screen-shot-2013-08-19-at-9.36.38-AM.png
Screen-shot-2013-08-19-at-9.36.38-AM.png [ 13.65 KiB | Viewed 5054 times ]
** use a calculator! ** (you wouldn’t be required to do this kind of math on the real test)

There are, therefore, five unique pairs of {a, b} values for which the equation a^2-b^2=225 is true.

The correct answer is (C).
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For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post Updated on: 01 May 2015, 10:01
2
1
(a+b)(a-b)=225

225 can be written as 5*5*3*3*1

so 225 can be written as product of two factors in 5 ways
5*45
25*9
225*1
15*15
75*3

solving above three equations, we get Five non negative pairs of (a,b)

Hence C

Originally posted by vijay556 on 01 May 2015, 08:06.
Last edited by vijay556 on 01 May 2015, 10:01, edited 1 time in total.
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Re: For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 01 May 2015, 09:35
2
225,1---1
15,0--2
45,9--3
75,3--4
25,9--5

hence ans c
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Re: For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 01 May 2015, 10:09
my take option A. 15,0. Have to wait another couple of days for OA.



Bunuel wrote:
For how many unique pairs of nonnegative integers {a, b} is the equation a^2 - b^2 = 225 true?

A) 1
B) 3
C) 5
D) 7
E) 9


Kudos for a correct solution.

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Kudos to you, for helping me with some KUDOS.
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Re: For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 01 May 2015, 10:38
2
Bunuel wrote:
For how many unique pairs of nonnegative integers {a, b} is the equation a^2 - b^2 = 225 true?

A) 1
B) 3
C) 5
D) 7
E) 9


Kudos for a correct solution.


Answer C
(a+b)(a-b)=225

5cases For (a+b), (a-b)

225, 1
75, 3
45, 5
25,20
15, 15



Answer 5
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Re: For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 01 May 2015, 10:47
3
1
225 = 1*3*5*3*5 = (a+b)*(a-b)
Its worth mentioning that if a - b > a + b, then b < 0 which doesn't work so we need to pay attention to that
Options:
a+b = 3*3*5*5, a-b = 1
a+b = 3*3*5, a-b = 5
a+b = 3*5*5, a-b = 3
a+b = 3*5, a-b = 3*5
a+b = 5*5, a-b = 3*3
rest of the options don't work coz a-b ends up bigger than a+b

C is the answer
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For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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New post 20 Feb 2016, 11:41
Bunuel wrote:
For how many unique pairs of nonnegative integers {a, b} is the equation a^2 - b^2 = 225 true?

A) 1
B) 3
C) 5
D) 7
E) 9


Kudos for a correct solution.



almost got me tricked this question...
ok, so a^2-b^2=225
we can re-write it as:
(a+b)(a-b)=225 or 15^2.
15^2 can be rewritten as 3^2 * 5^2. we thus have 9 factors.
let's see:
225; 1
75; 3
45; 5
9; 25
15; 15

since all are negative, for each "pair" we can find unique values for a and b to satisfy the conditions:
a+b=225; a-b=1
a+b=75; a-b=3
a+b=45; a-b=5
a+b=25; a-b=9
a+b=15; a-b=15

since we are asked only non-negative numbers, we can suppose that one of the numbers is 0.
thus, we see that we have 5 such pairs.
but all this is too time consuming, is there another approach?
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Re: For how many unique pairs of nonnegative integers {a, b} is the equati  [#permalink]

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Re: For how many unique pairs of nonnegative integers {a, b} is the equati   [#permalink] 11 Jul 2019, 04:03
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