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i, j, and k are non-negative integers such that i+j+k=3. If p, q, and

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i, j, and k are non-negative integers such that i+j+k=3. If p, q, and [#permalink]

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[GMAT math practice question]

\(i, j\), and \(k\) are non-negative integers such that \(i+j+k=3\). If \(p, q\), and \(r\) are three fixed, but different, prime numbers, how many different values of \(p^iq^jr^k\) are possible?

A. 8
B. 9
C. 10
D. 11
E. 12
[Reveal] Spoiler: OA

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Re: i, j, and k are non-negative integers such that i+j+k=3. If p, q, and [#permalink]

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New post 22 Nov 2017, 02:57
MathRevolution wrote:
[GMAT math practice question]

\(i, j\), and \(k\) are non-negative integers such that \(i+j+k=3\). If \(p, q\), and \(r\) are three fixed, but different, prime numbers, how many different values of \(p^iq^jr^k\) are possible?

A. 8
B. 9
C. 10
D. 11
E. 12


Hi...

Since I,j,k have a sum of 3, we have to find ways it is possible..
1) all three are 1..... One way
2) 0,1,2.....…............. 3! Ways=6 ways
3) 0,0,3….……….......... 3!/2!=3 ways

Total = 1+6+3=10

C
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i, j, and k are non-negative integers such that i+j+k=3. If p, q, and [#permalink]

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New post 22 Nov 2017, 03:58
The question basically asks how many whole number solution is possible
here the answer is 5c2 = 10 ways
therefore the answer must be 10
DIRECT FORMULA : (n+r-1) C (r-1)
here n= 3 and r=3

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Re: i, j, and k are non-negative integers such that i+j+k=3. If p, q, and [#permalink]

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The number of possible values of \(p^iq^jr^k\) is equal to the number of solutions of the equation \(i + j + k = 3\).
The solution set of the equation \(i + j + k = 3\) includes all permutations of \((3,0,0), (2,1,0), and (1,1,1)\).
The number of permutations of \((3,0,0)\) is \(\frac{3!}{2!} = 3\).
The number of permutations of \((2,1,0)\) is \(3! = 6.\)
The number of permutations of \((1,1,1)\) is \(1\).

Therefore, the number of solutions of the equation \(i+j+k=3\) is \(3 + 6 + 1 = 10\).

Therefore, the answer is C.

Answer: C
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Re: i, j, and k are non-negative integers such that i+j+k=3. If p, q, and [#permalink]

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New post 27 Nov 2017, 12:32
MathRevolution wrote:

\(i, j\), and \(k\) are non-negative integers such that \(i+j+k=3\). If \(p, q\), and \(r\) are three fixed, but different, prime numbers, how many different values of \(p^iq^jr^k\) are possible?

A. 8
B. 9
C. 10
D. 11
E. 12


Let’s first determine the number of non-negative integer triples (i, j, k) we can have:

(0,0,3), (0,3,0), (3,0,0)
(0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), (2,1,0)
(1,1,1)

We see that there are 10 triples, and when they are assigned as the exponents for p, q and r, we will have 10 different values of p^i * q^j * r^k.

Answer: C
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Re: i, j, and k are non-negative integers such that i+j+k=3. If p, q, and   [#permalink] 27 Nov 2017, 12:32
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