@Neha Kochhar • 17 Jan, 2010

What is the value of the expression (3^6) + (a^a)?

a) 3 b) 5 c) 6 d) a+18 e) None

I am Unable to solve this Out?

Ans is 5 but how?

Explain the Process

a) 3 b) 5 c) 6 d) a+18 e) None

I am Unable to solve this Out?

Ans is 5 but how?

Explain the Process

@Kaustubh Katdare • 17 Jan, 2010
Is this a C programming question?

@Manish Goyal • 17 Jan, 2010

ok

If you don't input the value of a then it will take garbage value..now in order to obtain output 5 in the expression it should be like this

(3^6)+(a^-a)

ie a's power is a negative value

I think you want to solve this expression using c right?Neha KochharWhat is the value of the expression (3^6) + (a^a)?

a) 3 b) 5 c) 6 d) a+18 e) None

I am Unable to solve this Out?

Ans is 5 but how?

Explain the Process

ok

If you don't input the value of a then it will take garbage value..now in order to obtain output 5 in the expression it should be like this

(3^6)+(a^-a)

ie a's power is a negative value

@Neha Kochhar • 21 Jan, 2010

Please explain it .I did not get what you said

but how the value 5 is obtained?goyal420I think you want to solve this expression using c right?

ok

If you don't input the value of a then it will take garbage value..now in order to obtain output 5 in the expression it should be like this

(3^6)+(a^-a)

ie a's power is a negative value

Please explain it .I did not get what you said

@Hussanal Faroke • 25 Jan, 2010
is the 'a' defined, if no try it in another system? or u defined pls tell the value of a! u know in c 32767+1 is -32765

@rishita • 11 Jun, 2011
Nice Goyal. But, we want detail solution. How we get the answer. Can you give detail step by step process of solution.

Thanks for your solution.

Regards,

<link removed>

Thanks for your solution.

Regards,

<link removed>

@vik001ind • 11 Jun, 2011
It doesn't matter whether a is defined or not.

^ stands for XOR. Xoring same bit return 0, different values return 1. Eg, 1^0 == 1 , 1^1 == 0

Any int variable in C is 16 bit (16 bit compiler) or 32 bit (32 bit compiler). So, in any case whether it is defined or not, a will be a 16/32 bit pattern.

Considering 16 bit compiler

Bit pattern of 3 is 0000 0000 0000 0000 0011

XOR

Bit pattern of 6 is 0000 0000 0000 0000 0110

Result is --> 0000 0000 0000 0000 0101 ---> 5

a^a will always be equal to 0. Since we have bit pattern same in both cases.

Therefore (3^6) + (a^a) = 5.

^ stands for XOR. Xoring same bit return 0, different values return 1. Eg, 1^0 == 1 , 1^1 == 0

Any int variable in C is 16 bit (16 bit compiler) or 32 bit (32 bit compiler). So, in any case whether it is defined or not, a will be a 16/32 bit pattern.

Considering 16 bit compiler

Bit pattern of 3 is 0000 0000 0000 0000 0011

XOR

Bit pattern of 6 is 0000 0000 0000 0000 0110

Result is --> 0000 0000 0000 0000 0101 ---> 5

a^a will always be equal to 0. Since we have bit pattern same in both cases.

Therefore (3^6) + (a^a) = 5.

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