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Bunuel
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If a, b and c are distinct positive integers, what is the value of (a 06 Aug 2017, 00:57
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95% (hard)

Question Stats:

37% (03:03) correct 63% (02:40) wrong based on 191 sessions

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If a, b and c are distinct positive integers, what is the value of (a + b + c)?

(1) \(2^{(2a+c)} + 3^b = 91\)

(2) \(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)
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A
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If a, b and c are distinct positive integers, what is the value of (a 06 Aug 2017, 04:24
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Bunuel
If a, b and c are distinct positive integers, what is the value of (a + b + c)?

(1) \(2^{(2a+c)} + 3^b = 91\)

(2) \(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)

(1)
\(2^{(2a+c)} + 3^b = 91\)
91 can only be 64 + 27
2a+c = 6 and b=3
now (a,b,c) can only be (1,3,4) (distinct positive integers)

Sufficient

(2)
\(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)
13 can only be 4+9
c/4 = 2
c=8
a+2b = 2 (not possible) cannot identify the value of a+b as b or a cannot be 0
Not sufficient

A
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If a, b and c are distinct positive integers, what is the value of (a 07 Nov 2017, 13:29
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Hi Bunuel, how come statement 2 is valid given a,b,c distinct positive integers?

Please clarify

thanks
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If a, b and c are distinct positive integers, what is the value of (a 07 Nov 2017, 13:46
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hellosanthosh2k2
Hi Bunuel, how come statement 2 is valid given a,b,c distinct positive integers?

Please clarify

thanks

Hey hellosanthosh2k2

B is not valid, that is the reason the answer is Option A.
If you have some other problem do let me know!

As already explained, a+2b = 2 is possible in 2 cases
case 1
a=0,b=1
case 2
b=0,a=2
As c has the value of 8, the value of the expression a+b+c could be 9 or 10

Hence, Option B is not sufficient
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If a, b and c are distinct positive integers, what is the value of (a 23 Dec 2017, 14:01
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it seems that there is no faster way to calculate. It tooks me 3.5 min. Also, the assumption here is that the value of (a+b+c) must have a value. In other words, if there is no value, then the statement is insufficient. For this reason, B is wrong.
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If a, b and c are distinct positive integers, what is the value of (a 09 Mar 2019, 02:42
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Luckisnoexcuse
Bunuel
If a, b and c are distinct positive integers, what is the value of (a + b + c)?

(1) \(2^{(2a+c)} + 3^b = 91\)

(2) \(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)

(1)
\(2^{(2a+c)} + 3^b = 91\)
91 can only be 64 + 27
2a+c = 6 and b=3
now (a,b,c) can only be (1,3,4) (distinct positive integers)

Sufficient

(2)
\(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)
13 can only be 4+9
c/4 = 2
c=8
a+2b = 2 (not possible) cannot identify the value of a+b as b or a cannot be 0
Not sufficient

A


In my opinion even A is not sufficient.

in A we get 2A+C=6

We get two possible solutions to the above ie. A=1,C=4 or A=2, C=2.

With two possible solutions, we get two different values of A+B+C

Bunuel, please help
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If a, b and c are distinct positive integers, what is the value of (a 07 Feb 2020, 06:01
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Luckisnoexcuse
Bunuel
If a, b and c are distinct positive integers, what is the value of (a + b + c)?

(1) \(2^{(2a+c)} + 3^b = 91\)

(2) \(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)

(1)
\(2^{(2a+c)} + 3^b = 91\)
91 can only be 64 + 27
2a+c = 6 and b=3
now (a,b,c) can only be (1,3,4) (distinct positive integers)

Sufficient

(2)
\(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)
13 can only be 4+9
c/4 = 2
c=8
a+2b = 2 (not possible) cannot identify the value of a+b as b or a cannot be 0
Not sufficient

A

Hi experts Bunuel VeritasKarishma ScottTargetTestPrep

Is there any way to trace the number patterns that way?

I find it quite complicated beyond certain limit. Please help!

Thanks.
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If a, b and c are distinct positive integers, what is the value of (a 10 Feb 2020, 02:13
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sahibr
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Bunuel
If a, b and c are distinct positive integers, what is the value of (a + b + c)?

(1) \(2^{(2a+c)} + 3^b = 91\)

(2) \(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)

(1)
\(2^{(2a+c)} + 3^b = 91\)
91 can only be 64 + 27
2a+c = 6 and b=3
now (a,b,c) can only be (1,3,4) (distinct positive integers)

Sufficient

(2)
\(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)
13 can only be 4+9
c/4 = 2
c=8
a+2b = 2 (not possible) cannot identify the value of a+b as b or a cannot be 0
Not sufficient

A


In my opinion even A is not sufficient.

in A we get 2A+C=6

We get two possible solutions to the above ie. A=1,C=4 or A=2, C=2.

With two possible solutions, we get two different values of A+B+C

Bunuel, please help

A and C are distinct integers. So A = 2, C = 2 is not allowed.
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If a, b and c are distinct positive integers, what is the value of (a 10 Feb 2020, 02:16
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Anurag06
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Bunuel
If a, b and c are distinct positive integers, what is the value of (a + b + c)?

(1) \(2^{(2a+c)} + 3^b = 91\)

(2) \(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)

(1)
\(2^{(2a+c)} + 3^b = 91\)
91 can only be 64 + 27
2a+c = 6 and b=3
now (a,b,c) can only be (1,3,4) (distinct positive integers)

Sufficient

(2)
\(2^{(a+2b)} + 3^{(\frac{c}{4})} = 13\)
13 can only be 4+9
c/4 = 2
c=8
a+2b = 2 (not possible) cannot identify the value of a+b as b or a cannot be 0
Not sufficient

A

Hi experts Bunuel VeritasKarishma ScottTargetTestPrep

Is there any way to trace the number patterns that way?

I find it quite complicated beyond certain limit. Please help!

Thanks.

There aren't too many options to worry about here.
\(2^{(2a+c)} + 3^b = 91\)
The only powers of 3 less than 91 are 3, 9, 27 and 81.
When subtracted from 91, they should leave a power of 2. You can arrive at 27-64 combination fairly quickly.
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If a, b and c are distinct positive integers, what is the value of (a 10 Feb 2020, 17:55
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Anurag06

Is there any way to trace the number patterns that way?

I find it quite complicated beyond certain limit. Please help!

Thanks.

In either case, you can subtract consecutive powers of one base and see if you get a power of the other base. For instance, to find powers of 2 and 3 that add up to 91, let's subtract powers of 3 from 91:

91 - 3^0 = 90
91 - 3^1 = 87
91 - 3^2 = 82
91 - 3^3 = 64 (a power of 2)
91 - 3^4 = 10
91 - 3^5 = a negative number

We see that the only way to add a positive integer power of 2 and 3 and get 91 is 3^3 + 2^6 = 91.
For powers of 2 and 3 that add up to 13, let's do the same but now subtract powers of 2:

13 - 2^0 = 12
13 - 2^1 = 11
13 - 2^2 = 9 (a power of 3)
13 - 2^3 = 5
13 - 2^4 = a negative number

Again we see that the only positive integer powers of 2 and 3 that add up to 13 is 2^2 + 3^2 = 13.
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