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How many different pairs of integers (a, b) exist such that 2 <= a <= 200 and a + b > a^b? (Two pairs of numbers are considered different if either a or b differs. For example, (2, 3) and (2, 4) are considered different, although they don’t satisfy the requirements of this problem.)

How many different pairs of integers (a, b) exist such that 2 <= a <= [#permalink]

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04 May 2015, 08:24

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\(a + b > a^b\) \(2 <= a <= 200\) which means a is positive Easiest way is to try plug numbers 1)\(b = 0\) a > 1 - good, so the amount of pairs we get from this is \(200 - 2 + 1 = 199\) 2)\(b = 1\) \(a + 1 > a => 1 > 0\), good, 199 pairs again 3)\(b = 2\) \(a + 2 > a^2\) \(a^2 - a - 2 < 0\) Since a is an integer and is between 2 and 200, and \(a^2 - a\) only grows when we increase a (coz \(a^2 > a\) if \(a > 1\)) we need to check what happens at the least possible value of a: when \(a = 2\) the result is \(4 - 2 - 2 = 0\) so when \(a > 2\), the result will be greater than 0 thus not be good for the inequality, this way we get 0 pairs. 4)\(b = 3\) \(a^3 - a - 3 < 0\) same story as above but in this case when we put 2, we are even further away from 0: 8 - 2 - 3 =5 < 0, at 3 and higher we will just increase the distance from 0, thus 0 pairs again It is fairly obvious now that at higher values of b we get 0 pairs so I guess that its for positive values of b. 5)\(b = -1\) \(a - 1 > \frac{1}{a}\) multiply both sides by a (coz a is positive) \(a^2 - a - 1 > 0\) a = 2: \(4 - 2 - 1 = 1 > 0\), good increasing a will give us even bigger results, so its safe to assume that this is 199 pairs again 6)\(b = -2\) \(a - 2 > \frac{1}{a^2}\) \(a^3 - 2*a^2 - 1 > 0\) \(a^2*(a - 2) - 1 > 0\) \(a^2*(a-2)\) becomes 0 when a = 2 and thus the whole experssion turns into -1 which is not bigger than 0 - fail, on the other hand if a > 2, the experssion becomes much bigger than 1 so our inequality holds, this way we can exclude just 1 value of a from our total, thus we get 198 pairs 7)\(b = -3\) \(a - 3 > \frac{1}{a^3}\) \(a^4 - 3*a^3 - 1 > 0\) \(a^3*(a - 3) - 1 > 0\) same story as above with only difference that a > 3, otherwise left side is definitely negative so we exclude 2 values of a this time (2 and 3), thus we are left with 197 pairs Its safe to assume that going further we will just shrink our total available values of a by 1 each time we decrease our b, so at some point we will be left with 0 values of a available. Since we always change b, all these pairs count so the resulting summ will be as follows: b = 1: 199 b = 0: 199 b = -1: 199 b = -2: 198 b = -3: 197 ... b = -199: 1 so the resulting summ is \((1 + 2 + ... 199) + 2*199 = 199*(1 + 199)/2 + 2*199 = 102*199 = 20298\)

How many different pairs of integers (a, b) exist such that 2 <= a <= [#permalink]

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04 May 2015, 09:30

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Bunuel wrote:

How many different pairs of integers (a, b) exist such that 2 <= a <= 200 and a + b > a^b? (Two pairs of numbers are considered different if either a or b differs. For example, (2, 3) and (2, 4) are considered different, although they don’t satisfy the requirements of this problem.)

A) 594 B) 738 C) 5,940 D) 19,800 E) 20,298

Kudos for a correct solution.

Given : a is +ive and 2 <= a <= 200 a + b > a^b

Solution :

if a=2 then \(2+b >2^b\) , possible values of 'b' are -1,0,1 , so there will be 3 pairs of (a,b) if a=3 then \(3+b >3^b\) , possible values of 'b' are -2,-1,0,1 , so there will be 4 pairs of (a,b) if a=4 then \(4+b >4^b\) , possible values of 'b' are -3,-2,-1,0,1 , so there will be 5 pairs of (a,b)

likewise if a=200, there will be 201 (a,b) pairs.

total possible pairs \(= 3+4+5+....+201 = \frac{(201+3)}{2} * 199= 20298\).

Answer E _________________

Thanks, Lucky

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How many different pairs of integers (a, b) exist such that 2 <= a <= 200 and a + b > a^b? (Two pairs of numbers are considered different if either a or b differs. For example, (2, 3) and (2, 4) are considered different, although they don’t satisfy the requirements of this problem.)

A glance at the answer choices shows that you’re not going to be counting out all of the individual pairs, so there must be some kind of pattern. How can you find it?

A restriction is given on a, but not on b (except that it’s an integer), so it’s best to begin the reasoning with a.

Try a = 2. The equation becomes 2+b>2^b. How to solve that? Think theoretically just a little bit. In general, raising 2 to a power will result in a number that grows more quickly than merely adding 2 to something. What are a few circumstances in which the 2+b portion will actually be bigger? You would need values for b that would minimize 2^b.

b=0 would work: 2+0>2^0. So would b=1: 2+1>2^1. If you go any bigger than that though, then the 2b side will get too big (e.g. if b=2, then 2+2=2^2). What about the other direction: negative integers? If b=-1, then 2+(-1)<2^(-2). Remember that a negative exponent will not cause a positive integer base to become negative. Rather, the base will just become a positive fraction.) There are three possible b values then when a=2, so there are 3 pairs of integers here that fulfill the inequality.

Okay, so what about a=3? Using the same “testing cases” method shown above, 3+b>3^b is true for four values: b=-2,-1,0, 1.

Try a=4. This time 4+b>4^b, which is true for five values: b=-3,-2,-1,0, 1.

Hmm. So far:

When a=2, b has 3 values: -1,0,1.

When a=3, b has 4 values: -2,-1,0,1.

When a=2, b has 3 values: -3,-2,-1,0,1.

That seems like a pattern. Each time, b hasn’t been able to be larger than 1. Each time, b had added one more negative integer. Does it hold all the way to a=200?

If you raise 2 to an integer larger than 1, it’s going to start getting bigger very rapidly, much more rapidly than adding that same number to 2. It makes sense that b can never be larger than 1. Okay, so that side of the pattern holds.

What about the negative end? Especially, the a+b side needs to stay positive in order to be greater than the a^b side, because the a^b side can never be zero or negative. In that case, if b is negative, the magnitude (distance from zero) has to be smaller than a‘s magnitude. If a=3, then b can’t be -3,-4,-5, and so on. If a=4, then b can’t be -4 or smaller. So this side of the pattern also holds.

The pattern also shows that the number of pairs for any value of a is equal to a+1. The number of possible pairs, then, is 3+4+5+...+201. (The final possible value for a is 200.)

There are 201-3+1=199 items with an average value of (3+201)/2=102, so the sum is 199*102=20298.

Re: How many different pairs of integers (a, b) exist such that 2 <= a <= [#permalink]

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28 May 2017, 04:23

Hi Bunuel, How much time do you think one should spend on a question like this in the actual exam? It took me around 3 minutes before I could observe any pattern so I gave up because it wasn't worth then. I had solved for b=0,-1,1,2 till then. Should I have stretched some more?

Hi Bunuel, How much time do you think one should spend on a question like this in the actual exam? It took me around 3 minutes before I could observe any pattern so I gave up because it wasn't worth then. I had solved for b=0,-1,1,2 till then. Should I have stretched some more?

This is a Manhattan GMAT challenge problem. If you get something that hard on the exam it would probably mean that you are already doing very well, which would also mean that you are good at math and saved some time on easier questions. If all this is true, then at this stage you'd have several spare minutes accumulated and you'll be able to spend 1-2 minutes more on this question (again if you feel confident enough). You can see in the stats above that the average time of correct answer is 03:47 minutes, so I'd say that it's possible to solve it in 3 minutes.
_________________

How many different pairs of integers (a, b) exist such that 2 <= a <= [#permalink]

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21 Sep 2017, 01:59

Here is my take.

I first started with non-negative integers. For example, I tried forming the inequality using the integer 2 first. I realized that there were 2 cases, when a=2 and b=0. The other case is when a=2 and b= 1. So we have two cases for the integer 2. Then I moved to the integer 3, here as well you will find two cases- one in which a=3 and b=0 and the other in which a-3 and b=1. On further speculation you will realize that there are 2 possible cases for each of the values of a from 2 to 100. How many cases are there? Lets calculate: 2 to 200= 199 integer values of a. For each of these, we have two possible cases. So total number of cases=398

NOW LETS SEE IF ANY NEGATIVE VALUES ARE POSSIBLE FOR B Lets start with a =2, now you will realize that there is only one such possible case- One in which a=2 and b=-1. Come to a=3, 2 possible cases- a=3, b=-1 and a=3 and b= -2 Come to a=4, 3 possible cases:-a=4, b=-1 | a=4, b=-2 and a=4 and b=-3

So there is a pattern. For every integral value of a from 2 to 200, there are (n-1) cases, when n refer to the integer being chosen.

So we have 1+2+3+4+......+199 = 199(100)=19,900

Now adding up the total cases: 19,900+ 398 = 20298

Answer is E
_________________

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