What is the greatest integer less than or equal to (3^100 + 2^100)

Question

What is the greatest integer less than or equal to  \frac{3^{100}+2^{100}}{3^{96}+2^{96}} ?

(A) 80
(B) 81
(C) 96
(D) 97
(E) 625

nick1816 · Accepted Answer

\frac{3^{100}+2^{100}}{3^{96}+2^{96}} = {2^100( 1.5^100 +1)}/ {2^96(1.5^96)+1}

= 2^4 {\frac{1.5^{100} +1}{(1.5^{96}+1)}}

2^4 {\frac{( 1.5^{100} +1)}{(1.5^{96}+1)}} is slightly less than  16*(\frac{1.5^{100}}{1.5^{96}}) =  16*(1.5^4) =16*(81/16)=81

the greatest integer less than or equal to  \frac{3^{100}+2^{100}}{3^{96}+2^{96}} =80

BrentGMATPrepNow · Answer

The key here is to recognize that,   compared to 3^100, the value of 2^100 is negligible.

To see why, let's examine some powers of 2 and 3

2^3 = 8 and 3^3 = 27
2^5 = 32 and 3^5 = 243
2^7 = 128 and 3^7 = 2187
2^9 = 512 and 3^9 = 19,683
2^11 = 2048 and 3^11 = 177,147
2^20 = 1,048,576 and 3^20 = 3,486,784,401

As you can imagine 3^100 will be so much bigger than 2^100, that the value of 2^100 has little effect on the sum 3^100 + 2^100
The same can be said of the effect that 2^96 has on the sum 3^96 + 2^96

So.... \frac{3^{100}+2^{100}}{3^{96}+2^{96}}≈\frac{3^{100}}{3^{96}}≈3^4≈81

So, the correct answer is APPROXIMATELY 81, which means as can ELIMINATE C, D and E

The ORIGINAL expression is  \frac{3^{100}+2^{100}}{3^{96}+2^{96}}

Since we are adding more to the numerator (we're adding 2^100) than we are adding more to the denominator (we're adding 2^96), we can conclude that  \frac{3^{100}+2^{100}}{3^{96}+2^{96}}  is greater than  \frac{3^{100}}{3^{96}}

In other words,  \frac{3^{100}+2^{100}}{3^{96}+2^{96}}  is greater than 81

Answer: B

ScottTargetTestPrep · Answer

Since 2^100 is actually a lot less than 3^100 and similarly, 2^96 is also a lot less than 3^96, so the given quotient is approximately 3^100/3^96 = 3^4 = 81. So the answer could be 81 or 80. To determine which one it should be, we can see if (3^100 + 2^100)/(3^96 + 2^96) ≤ 80:

(3^100 + 2^100)/(3^96 + 2^96) < 3^4 ?

3^100 + 2^100 < 3^4(3^96 + 2^96) ?

3^100 + 2^100 < 3^100 + 3^4 x 2^96 ?

2^100 < 3^4 x 2^96 ?

2^4 x 2^96 < 3^4 x 2^96 ?

2^4 < 3^4 ? → Yes!

Since (3^100 + 2^100)/(3^96 + 2^96) is strictly less than 3^4 = 81, therefore, the greatest integer less than or equal to (3^100 + 2^100)/(3^96 + 2^96) is 80.

Answer: A

sebastianm · Answer

I tried to solve the question using a more exam-friendly approach, let me know if this method has some flaws.

(3^100+2^100)/(3^96+2^96) =

(3^4*3^96+2^4*3^96)/(3^96+2^96)

(81*3^96+16*3^96)/(3^96+2^96)

at this point, it is impossible any kind of simplification so I need to rearrange the terms in this way

(81*3^96+81*3^96 - 65*3^96)/(3^96+2^96) =

(81*3^96+81*3^96)/(3^96+2^96) - (65*3^96)/(3^96+2^96) =

81 - (65*3^96)/(3^96+2^96)

At this point, we see that the fraction is equal to 81 less something very small (65*3^96)/(3^96+2^96) and so we can conclude that the greater integer less than the fraction is 80 and therefore the  answer is A

Leave Kudos if you like this method!

puneetb · Answer

===3^4 because:::Num

Abir77 · Answer

I took a different approach. I saw that options have different unit value. So i calculated unit value of 3^100 = 1, 2^100 =6, 3^96 = 1, 2^96 = 6
That means 
 \frac{3^{100}+2^{100}}{3^{96}+2^{96}}  = (1+6)/(1+6) = 7/7 
That means unit value of ans will have 0. So ans is 80
 Bunuel  please tell me the flaws

sghosh1096 · Answer

let 3^96=k and 2^96=m

the expression becomes

(k*3^4 + m*2^4)/(k+m) = (81*k+16*m)/(k+m) = 81 - 65*m/(m+k)

now 65m < k+m

so the number be something like 81 - 0.abcd
therefore the greatest integer less than or equal to the expression is  80 .... (A)

What is the greatest integer less than or equal to (3^100 + 2^100)

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What is the greatest integer less than or equal to (3^100 + 2^100)

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