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Some Nice integer questions

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Some Nice integer questions

by deepoe » Sun Mar 01, 2009 12:06 pm
1. Let a, b, and c be three integers, and let a be a perfect square. If a/b = b/c, then which one of
the following statements must be true?
(A) c must be an even number
(B) c must be an odd number
(C) c must be a perfect square
(D) c must not be a perfect square
(E) c must be a prime number

OA=C

2. If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following
formula: S = n(n + 1)/2. Which one of the following statements about S must be true?
(A) S is always odd.
(B) S is always even.
(C) S must be a prime number.
(D) S must not be a prime number.
(E) S must be a perfect square.

OA=D

3. A number, when divided by 12, gives a remainder of 7. If the same number is divided by 6,
then the remainder must be
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
OA=A

4. Let x be a two-digit number. If the sum of the digits of x is 9, then the sum of the digits of the
number (x + 10) is
(A) 1 (B) 8 (C) 10 (D) either 8 or 10 (E) either 1 or 10

OA=E
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by DanaJ » Sun Mar 01, 2009 1:05 pm
Q1.
Consider a/b = b/c. This means that a = (b^2)*c. Now, since a is a perfect square and b^2 is obviously a perfect square, then c, which is a/(b^2) must also be a perfect square.

Q2.
S being an odd or an even number depends on the nature of n. As you may know, n(n + 1) is always even (the product of two consecutive numbers is always even), but when you divide it by 2, you may have either an odd or an even number. Examples:
n = 4 makes S = 10 (even)
n = 5 makes S = 15 (odd)
S being a prime number is the easiest to eliminate. The two examples above are more than sufficient for that.
S must not be a prime number is the correct choice here and this happens because n(n + 1) is even. When you divide it by 2, you have two cases:
a. n is even - this makes n/2 an integer and, since n+1 is also an integer, you have that S is the product of two integers
b. n + 1 is even - this makes (n + 1)/2 an integer and n is also an integer, and you follow the same reasoning as above.
E can be again easily eliminated with the two examples above.

Q3.
Say x = your number. Then you have that x = 12q + 7 = 6*2*q + 6 + 1 = 6(2q + 1) + 1. This is why the remainder will be indeed 1 when you divide x by b.

Q4. x has the following format: MN, with M, N digits from 1 to 9 (N can also be 0). The tens position of MN + 10 will be M + 1, since you added 10. An example would be:
MN = 34
MN + 10 = 44 - the tens position is (initial position + 1)
This means that the new sum of digits will be M + N + 1 = (M + N) + 1 = 10.
But we must not forget of the special case when M is 9. In this case, MN + 10 will have the "10N" format. Example:
MN = 94
MN + 10 = 104.
This makes the sum of digits N + 1. Now, since M is considered 9 in this special case, this means that from M + N = 9 you get that N = 0. This makes the sum of digits for the special case 1.
So the answer will be 10 or 1.
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by deepoe » Sun Mar 01, 2009 1:33 pm
I don't understand Q4 really:$

But damnnn you're good hahaha, got some tips for me to be as good as you :oops:
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by cramya » Sun Mar 01, 2009 2:37 pm
Q4.

I think the examples given for 4 may be a little off or I am missing something which is quite possible

MN or x whatever-> the sum of the digits of this 2 digit number must be 9

Eg:

54
36
27

Add 10 to these and the number u get will have digit sum equalt to 10


Take 90 for example

90+10 = 100 sum of the digits of 100 is 1

Hence 1 or 10
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by DanaJ » Sun Mar 01, 2009 9:40 pm
I chose those examples without taking into consideration the fact that the sum of digits was 9, but merely to illustrate what happens when you add 10 to a two-digit number. Maybe that's what was a bit confusing.
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by cramya » Sun Mar 01, 2009 9:42 pm
Got u ;all's well!

Good luck, DanaJ.

Regards,
Cramya
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by welcome » Mon Mar 02, 2009 1:10 am
For Q4. Only 2 conditions are possible...

Condition 1 when no. comes with '0', say 90 then 90+10 = 100 => digit sum =1

Condition 2 when no. comes without '0', say 45 then 45+10 = 55 => digit sum =10

Answer E.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590
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by deepoe » Mon Mar 02, 2009 2:21 am
Ah thanks thanks :D:D
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by farooq » Mon Mar 02, 2009 2:51 am
Question 4. is little bit tricky :)

Good post.
Regards,
Farooq Farooqui.
London. UK

It is your Attitude, not your Aptitude, that determines your Altitude.
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