M37-87

Question

What is the coefficient of  m^9  if you expand the following product:  (1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111}) ?

A.  7 
B.  8 
C.  9 
D.  10 
E.  11

Bunuel · Answer

Official Solution:

What is the coefficient of  m^9  if you expand the following product:  (1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111}) ?

A.  7 
B.  8 
C.  9 
D.  10 
E.  11

When you expand  (1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{111})  you'll get:
 
  1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9+... .
 
We need to find the coefficient of  m^9 ,  i .

Notice that all terms where the exponent is higher that 9 ( (1 + m^{10}), \ (1 + m^{11}), \ ..., \ (1 + m^{111}) ) will not contribute to the coefficient of  m^9  so they can be ignored. For example, say we have:  (1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^{9})(1 + m^{10}) .
 
When we expand the first nine terms we'd get: 
 
 (1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)(1 + m^{10}) 
 
And when we further expand this we'd get: 
 
 1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9+ 
 
 + (1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)*m^{10} . 
 
Notice that the coefficient of  m^9 ,  i , remained the same (all terms of  (1 + am + bm^2+cm^3+dm^4+em^5+fm^6+gm^7+hm^8+im^9)*m^{10} ) will have  m  in higher power than 9).

So, we can focus only on  (1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^9) 
 
Now, we know that  a^x*a^y*a^z*...=a^{x+y+z+...} , so to get the coefficient of  m^9  we should find out in how many ways the exponents can sum to 9.
 
If we have one term we can just have  m^9 . 1 way.
 
If we have two terms we can have  m*m^8=m^9 ,  m^2*m^7=m^9 ,  m^3*m^6=m^9 , or  m^4*m^5=m^9 . 4 ways.
 
If we have three terms we can have  m*m^2*m^6=m^9 ,  m*m^3*m^5=m^9 , or  m^2*m^3*m^4=m^9 . 3 ways.
 
Four or more terms will not give the exponent of 9 because the least exponent will be  m*m^2*m^3*m^4=m^{10} .
 
Total of  1+3+4=8  ways.
 So, the coefficient of  m^9  is 8 For those who are interested: 
 (1 + m)(1 + m^2)(1 + m^3)(1 + m^4)...(1 + m^9)= 
 
 1 + m + m^2 + 2 m^3 + 2 m^4 + 3 m^5 + 4 m^6 + 5 m^7 + 6 m^8 + 8 m^9 +... + m^{45} .

Answer: B

Pawan01236 · Answer

Bunuel  
1. for two terms: How to know that there are only single terms of m^1 and m^8? (i.e. how it is not possible that 2m * 4 m^8 = 8m^9)
2. for three terms: How to know that m1*m4*m4 is not there?

Bunuel · Answer

(1 +  m )( 1  + m^2)( 1  + m^3)( 1  + m^4)( 1  + m^5)( 1  + m^6)( 1  + m^7)( 1  + m^8)( 1  + m^9)

When expanding the above we get m, only when we multiply m by eight 1's: m*1*1*1*1*1*1*1*1. No other way.

To get m^9 we can have:

1*1*1*1*1*1*1*1*m^9: 
 ( 1  + m)( 1  + m^2)( 1  + m^3)( 1  + m^4)( 1  + m^5)( 1  + m^6)( 1  + m^7)( 1  + m^8)(1 +  m^9 )

m*1*1*1*1*1*1*m^8*1:
 (1 +  m )( 1  + m^2)( 1  + m^3)( 1  + m^4)( 1  + m^5)( 1  + m^6)( 1  + m^7)(1 +  m^8 )( 1  + m^9)

1*m^2*1*1*1*1*m^7*1*1:
 ( 1  + m)(1 +  m^2 )( 1  + m^3)( 1  + m^4)( 1  + m^5)( 1  + m^6)(1 +  m^7 )( 1  + m^8)( 1  + m^9)

1*1*m^3*1*1*m^6*1*1*1 :
 ( 1  + m)( 1  + m^2)(1 +  m^3 )( 1  + m^4)( 1  + m^5)(1 +  m^6 )( 1  + m^7)( 1  + m^8)( 1  + m^9)

1*1*1*m^4*m^5*1*1*1*1:
 ( 1  + m)( 1  + m^2)( 1  + m^3)(1 +  m^4 )(1 +  m^5 )( 1  + m^6)( 1  + m^7)( 1  + m^8)( 1  + m^9)

m*m^2*1*1*1*m^6*1*1*1:
 (1 +  m )(1 +  m^2 )( 1  + m^3)( 1  + m^4)( 1  + m^5)(1 +  m^6 )( 1  + m^7)( 1  + m^8)( 1  + m^9)

m*1*m^3*1*m^5*1*1*1*1:
 (1 +  m )( 1  + m^2)(1 +  m^3 )( 1  + m^4)(1 +  m^5 )( 1  + m^6)( 1  + m^7)( 1  + m^8)( 1  + m^9)

1*m^2*m^3*m^4*1*1*1*1*1:
 ( 1  + m)(1 +  m^2 )(1 +  m^3 )(1 +  m^4 )( 1  + m^5)( 1  + m^6)( 1  + m^7)( 1  + m^8)( 1  + m^9)

Hope it's clear.

RJ_Joker · Answer

I like the solution - it’s helpful.  Bunuel  it is too long, can we expeact such kind of question in GMAT exam?

Bunuel · Answer

Yes, this question is hard, but writing out the full explanation takes much longer than actually solving it. Something like this could potentially come up on a real exam.

Dream009 · Answer

Bunuel - I don't understand this at all

Spent 45 mins racking my head. Also tried chatgpt. KarishmaB - any interesting short answer

Bunuel · Answer

You can check alternative explanations here: https://gmatclub.com/forum/hot-competit ... 32946.html Hope it helps.

KarishmaB · Answer

The question is a simple combinations problem. The major problem is how to arrive at the core combinations question:

Core Question: Given digits from 1 to 9, in how many ways can you make a sum of 9?

To solve it, you will simply enumerate it: 
Using only one digit: 9. So only 1 way
Using 2 digits: you can select any from 1 to 4 and its corresponding digit to make a 9 e.g. 1+ 8, 2+7 etc. So 4 ways
Using 3 digits: 1+2+6, 1+3+5, 2+3+4. So 3 ways
4 or more digits will give a sum of more than 9 so that is not possible. 
Hence 1+4+3 = 8 is the answer

But let's see how to arrive at this core question:

Take a smaller expression:  (1+m^1)(1+m^2)(1+m^3)(1+m^4) 
Multiply the factors together and see in how many ways you get m^3. 
 	ext{One Way}: 1*1*m^3*1 
 	ext{Another Way}:m^1*m^2*1*1 
These terms will add up to give 2m^3.
We can ignore higher powers because they will not give a power of 3. 
This understanding is how you arrive at the core combinations question. A power of 9 you can get by adding smaller exponents together. So the question boils down to - in how many ways can you add any number of digits from 1 to 9 to get a sum of 9. The number of ways will all add up to give the coefficient of m^9.

M37-87

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