Questions on integer

C

ctmckoay

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Oct 12, 2010
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13
Can someone help on this ? Appreciate your help.

(1) The average of 21 consecutive integers is 31. The largest of these integers must be ??

I've no clue as to how to derive the answer, which is A) 40, B)41, C)51 D)52 . Answer is B) 41.

Sorry I don't get how to derive the answer still after reading through your answer.

Are you saying..i should average -10,-9............10, i'm still lost, please help
 


Did somebody tell you that (C) is the answer?

If twenty-one consecutive integers average 31, the answer is (B), not (C).

However, if they average 41, instead, then (C) is the correct answer.

Please proofread your post, and correct any errors.

Also, have you yet learned the meaning of an average?

Since there are 21 consecutive numbers total, there must be 10 numbers on each side of the average, yes?

 
ctmckoay said:
Can someone help on this ? Appreciate your help.

(1) The average of 21 consecutive integers is 31. The largest of these integers must be ??

I've no clue as to how to derive the answer, which is A) 40, B)41, C)51 D)52 . Answer is B) 41.

Sorry I don't get how to derive the answer still after reading through your answer.

Are you saying..i should average -10,-9............10, i'm still lost, please help
Click to expand...
 
Further questions on integer

Hi,

I've got yet another question on integer where i don't know how to begin or for that matter do.

The number 1000 has only 16 positive whole number factors. The product of all 16 of these factors is ? The answer is 10 [sup:2fdaaf0e]24[/sup:2fdaaf0e]. Please help.
 
Factors come in pairs:

1, 1000
2, 500
4, 250
5, 200
8, 125
10, 100
20, 50
25, 40

Now multiply these all together.


Alternate method: Look at the prime factorization of 1000:

\(\displaystyle 1000=2^3\cdot 5^3\)

Now list all possible products of these factors:

\(\displaystyle 1, 5, 5^2,5^3, 2, 2*5,2*5^2,2*5^3,2^2,2^2*5, 2^2*5^2,2^2*5^3,2^3,2^3*5,2^3*5^2,2^3*5^3\)

Multiply by adding all the exponents of 5, and all the exponents of 2.
 
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