Download 3 If p is a positive odd integer, what is the remainder when p is

3 If p is a positive odd integer, what is the remainder when p is divided by 4?
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
(1) 足够回答问题，因为 p = 8q + 5。（q 是商。）
所以 p = 4(2q + 1) + 1。那么 p 被 4 除时余数是 1。
(2) 这个条件难很多。设 p = m^2 + n^2。因为 p 是奇数，所以 m 和 n 必须一奇一偶。设
m 是偶数，那么 m^2 就必须是 4 的倍数，所以可以不考虑。现在只要看 n^2。因为
n
是奇数，所以 n 可以写
Originally Posted by rits700
the positive integer k has exactly 2 positive prime factors 3 and 7. If k has a total of
6 positive factors including 1 & k, what is the value of k?
1) 3^2 is a factor of k
2) 7^2 is not a factor of k
If p is the smallest positive integer such that p^3/3920 is also an integer, what is the
sum of the digits of p?
(A) 5
(B) 7
(C) 9
(D) 11
(E) 13
Answer: A
m = p*p*p/ 3920
m = p*p*p/ 2*2*2*2*5*7*7
m = p*p*p/ [(4^2)*5*(7^2)]
Hence p must be equal to 4*5*7
thus p = 140
Hence sum of digits = 1+4+0 = 5
Posted by Prachi Pareekh at 3/19/2008 05:11:00 AM 0 comments Links to this post
Labels: Integers, Problem Solving
Tuesday, March 18, 2008
Problem Solving - 44
Samar tried to type his new 7-digit phone number on a form, but what appeared on
the form was 39269, since the '4' key on his computer no longer works. His
secretary has decided to make a list of all of the numbers that could be Samar's new
number. How many numbers will there be on the list?
(A) 21
(B) 24
(C) 25
(D) 30
(E) 36
Answer: A
It is clear from the question that there are two 4's missing as it is a seven digit
number.
Total number of ways of choosing places for these two missing 4's in these 7 digits is
7C2 = 21 Once you fix the place for these two 4's rest all numbers will occupy
remaining places.
Posted by Prachi Pareekh at 3/18/2008 08:51:00 AM 1 comments Links to this post
Labels: Permutations and Combinations, Problem Solving
Tuesday, February 05, 2008
Data Sufficiency - 44
In triangle ABC, AB has a length of 10 and D is the midpoint of AB. What is the
length of line segment DC?
(1) Angle C= 90
(2) Angle B= 45
Answer: A
From statement (1): it is given that angle C = 90 degrees ...this implies that ABC is a
right angle triangle with AB as the hypotenuse and DC as the median. We know
that --- In all right triangles, the median on the hypotenuse is the half of the
hypotenuse. Hence DC=5
Posted by Prachi Pareekh at 2/05/2008 12:41:00 AM 0 comments Links to this post
Labels: Data Sufficiency, Geometry
Sunday, February 03, 2008
Data Sufficiency - 43
In the figure shown, what is the value of x?
(1) The length of line segment of QR is equal to the length of line segment RS
(2) The length of line segment of ST is equal to the length of line segment TU
Answer: C
From statement (1): Length of line segment of QR is equal to the length of line
segment RS ..this implies angle RQS = angle RSQ = p(say)
From statement (2): Length of line segment of ST is equal to the length of line
segment TU .. this implies angle TUS = angle TSU = q(say)
Hence p+p+angle QRS = 180 --- eq(1) and q+q+angle UTS = 180 --- eq(2)
Thus, p+q+x = 180
Now because angle RPT = 90, QRS+UTS= 90
adding eq(1) and eq(2) we get:
2p+2q+QRS+UTS = 360
2p+2q+90=360
p+q = 270/2 = 135
Now x = 180-p-q..hence the answer C
x = 180 - (p+q) = 180 - 135 = 45
Posted by Prachi Pareekh at 2/03/2008 07:33:00 AM 1 comments Links to this post
Labels: Data Sufficiency, Geometry, GMAT Prep
Thursday, January 31, 2008
Data Sufficiency - 42
Is a-3b an even number?
1). b=3a+3
2). b-a is an odd number
Answer: C
From statement (1): Given that b=3a+3
Thus a-3b=a-3(3a+3) = -8a-9 which may be even, odd, integer, non-integer, rational
etc ... Hence insufficient
From statement (2): Given that b-a is an odd number implies b is of the form
b=(2k+1)+a where k is an integer
Thus a-3b= a-3[(2k+1)+a] = -2a -6k-3 which may be even, odd, integer, non-integer,
rational etc ..Hence insufficient
Taking statement (1) and (2) together: -8a-9=-2a-6k-3 for some integer k
or -6a=-6k+6=-6(k+1) implies a=k+1
Thus a is an integer, either odd or even
Now statement (2) tells us that b is also an integer and that exactly one of {a,b} is
even
If a is even and b is odd, a-3b is odd
If b is even and a is odd a-3b is odd
Thus (1) and (2) combined tell us that a-3b is an odd number...hence sufficient
Posted by Prachi Pareekh at 1/31/2008 11:18:00 PM 2 comments Links to this post
Labels: Data Sufficiency, Integers, Numbers
Wednesday, January 30, 2008
Problem Solving - 43
If the sum of four consecutive positive integers a three digit multiple of 50, the mean
of the these integers must be one of x possible values, where x=
(A) 7
(B) 8
(C) 9
(D) 10
(E) more than 10
Answer: C
Suppose four integers are a, a+1, a+2 and a+3
Hence a + (a+1) + (a+2) + (a+3) = 4a+6
Now 4a+6 will be a multiple of 50 i.e 4a+6=50m where m can take any value from
{2,3...,19}
But a = (50m-6)/4 = (25m-3)/2 must be an integer, so m must be odd.
Thus m can be any odd integer from 3 to 19
3=1+1*2
19=1+9*2
So there are 9 different values for m, a and 4a+6, as well as (4a+6)/4
Posted by Prachi Pareekh at 1/30/2008 06:25:00 PM 0 comments Links to this post
Labels: Integers, Problem Solving
Data Sufficiency - 41
Sania has a circular garden in her backyard. She puts poles A,B and C on the
circumference of her garden. Then she ties ropes between these poles. Is length of
one of the ropes is equal to the diameter of her garden?
1. Slope of line joining pole A and B is 3/4 and slope of line joining poles B and C is
-4/3
2. Length of line joining pole A and B is 12 and length of line joining B and C is 5
Answer: A
From statement (1): Given that the slope of line AB is 3/4 and slope of line BC is -4/3.
This implies that the product of slopes = -1. Hence AB perpendicular BC and B is a
right angle. Thus AC is a diameter which implies ABC form a semi-circle.
Hence sufficient
From statement (2): Given that length of line AB is 12 and length of BC is 5. However
this does not imply that ABC is a right angled triangle. We can draw number of
different triangles with the same given two sides but with different third side.
Hence insufficient
Posted by Prachi Pareekh at 1/30/2008 06:07:00 PM 0 comments Links to this post
Labels: Data Sufficiency, Geometry
Problem Solving - 42
E is a collection of four odd integers and the greatest difference between any two
integers in E is 4. The standard deviation of E must be one of how many numbers?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Answer: B
Suppose the integers are 1, 3 and 5. Therefore the four integers can be:
1, 5, 5, 5
1, 3, 5, 5
1, 3, 3, 5
1, 1, 5, 5
1, 1, 1, 5
1, 1, 3, 5
Here two pairs have the same standard deviation. thus in all we have four different
standard deviations
Posted by Prachi Pareekh at 1/30/2008 05:58:00 PM 0 comments Links to this post
Labels: Problem Solving, Statistics
Wednesday, January 23, 2008
Data Sufficiency - 40
In XY plane, does the line with equation y=3x+2 contain point (r,s)?
1) (3r + 2 - s)(4s + 9 - s) = 0
2) (4r - 6 - s)(3r + 2 - s) = 0
Answer: C
Given that y = 3x+2 implies that does 3x+2-y = 0 contains the point (r,s) implies is
(3r+2-s) = 0 ?
From statement (1): (3r+2-s)(4s+9-s) = 0 implies either (3r+2-s) = 0 or (4s+9-s) = 0.
Now when (3r+2-s)...the line passes through (r,s)
When (4r+9-s) = 0 ...we cannot determine that whether the line passes through (r,s)
or not.
Hence insufficient
From statement (2): (4r-6-s)(3r+2-s) = 0 implies either (4r-6-s) = 0 or (3r+2-s) = 0
Now when (4r-6-s) = 0 ... we cannot determine that whether the line passes through
(r,s) or not
When (3r+2-s) = 0..the line passes through (r,s)
Hence insufficient
Taking statement (1) and (2) together: (3r+2-s)(4s+9-s) = 0 and (4r-6-s)(3r+2-s)
=0... We cannot have both 4r+9-s=0 and 4r-6-s=0 so it is (3r+2-s) = 0 ... only this
equation makes both the equations to be 0
Hence sufficient