Manhattan Challenge Problem of the week ! july 24

Is the positive integer x odd?

(1) x = y2 + 4y + 6, where y is a positive integer.

(2) x = 9z2 + 7z - 10, where z is a positive integer.

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Answer - The correct answer is B.

For complete solution click on the link below.
Manhattan challenge problem

Maths - Prime Factors

Methods and tricks to solve questions related to Prime factors.

1). Counting the Number of Factors -- If you factor a number into its prime power factors, then the total number of factors is found by adding one to all the exponents and multiplying those results together.

Example: The total number of factors of 108 are --
108 = (2^2) * (3^3) thus we add 1 to the exponents and multiply the results together.

(2+1)*(3+1) = 3*4 = 12.

Hence total number of factors of 108 are 12.

Verifying the above

The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108 , which are total 12 in number.


2). Factoring Factorials - First write the number you are factoring as a product of two or more numbers. For example, suppose we want to factor the number 30.

We know that 30 = 5 x 6,

so we write for the first step of our factor tree:

30
/ \
5 6

Next we factor the factors, if possible. In other words, for each number in the product from the first step, we try to write it down as a product of even smaller numbers.

For instance, in our example, we would try to factor both 5 and 6. As we noted above, 5 is prime, so we can't factor it further.

We are done with that branch of the factor tree.

However, 6 is not prime. 6 = 2 x 3, so we can extend our factor tree as follows:

30
/ \
5 6

Now further

6
/ \
2 3

Now, we continue this process until all of the branches of the tree end in prime numbers.

In our example, we are done after two steps, since 5, 2, and 3 are all prime numbers.

The factors of the number are the numbers at the end of the different branches on the factor tree. To figure out what power each prime factor is raised to, count the number of times the prime factor appears in the factor tree.

In our example, each of the factors appears only once, so the prime factorization of 30 is: 30 = 2 x 3 x 5.

Problem Solving - 8

How many factors of 2940 are NOT factors of 112?

(A) 22

(B) 30

(C) 32

(D) 34

(E) None of these

Answer - B

Prime factors of 2940 = (2 ^ 2) * 3 * 5 * (7 ^ 7)

Hence total number of factors of 2940

= (2 + 1) * ( 1 + 1) * (1 + 1) * ( 2 + 1) = 36
(by the method of counting number of factors)

Prime factors of 112 = (2 ^ 4) * 7

Therefore common factors for the above two numbers are

2^2 * 7

Total number of factors for (2 ^ 2) * 7 = (2 + 1) * ( 1 + 1) = 6

Hence total factors of 2940 which are not factors of 112 are

36 - 6 = 30, Hence B is the answer.

Problem Solving - 7

Eight points lie on the circumfrence of a circle.What is the positive diff b/w th no. of triangles and the no. of quadrilaterals that can be formed by connecting these points?

a. 8

b. 14

c. 56

d. 70

e. 1,344

Answer - B

Since the points are on the circumference of the circle thus no 3 or 4 points will be colinear, thus all points can be used to draw a triangle or quadilateral

Hence 8C4 = 70 and 8C3 = 56 (we make use of combination here because here the order of points do not matter)

Taking the difference we get 70 - 56 = 14.

DS Question 13

Is the integer n odd?

(1) n is divisible by 3.

(2) 2n is divisible by twice as many positive integers as n.

Answer - B

Statement (1) - insufficient as n can be both even and odd.
Statement (2) - sufficient - Here n will have a unique prime factorisation as a product of

p1 ^ q1 * p2 ^ q2 * ....... pk ^qk

where p1 is less than p2 , p2 is less than p3 and so on.

Now p1 is either equal to 2 or it is an odd integer greater than 2.
Thus if n is odd then 2n has a unique prime factorisation of

2 * p1^q1 * p2^q2 * .... * pk ^ qk

but if n is even then 2n can be factorised as

p1^(q1+1) * p2^q2 * ... where p1=2, q1 >= 1

Hence , now the total number of factors is based on choice of prime powers for each prime in term.

Thus if n is odd, then 2n must have twice as many factors as n.

Alternatively in simpler language

If n is even --- 2n will not have twice as many divisors as n
e.g . For n=2; 2n or 4 has 1,2,4 as divisors ,
For n = 4 divisors are 1,2,4; 2n=8 has 1,2,4,8

But if n is odd
then if n=3 divisors are 1,3 and 2n=6 divisors are 1,2,3,6
n=5 divisors are 1,5 and 2n=10 divisors are 1,2,5,10

Thus if n is odd the 2nd condition is satisfied. Hence ans is B

Manhattan Challenge problem of the Week!

Dig those Digits - Challenge problem of the week

Click on the link below to see the question and its solution.
Manhattan challenge problem

DS Question - 12

Does x + y = xy?

(1) x is neither a positive integer nor a negative integer

(2) y is neither a positive integer nor a negative integer

Answer - E - x and y can be either 0 or fractions.. which makes both conditions insufficient.

Manhattan Challenge Problem of the Week - July 5

Louie takes out a three-month loan of $1000. The lender charges him 10% interest per month compunded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month?

(A) 333

(B) 383

(C) 402

(D) 433

(E) 483

Answer - C

For solution click on the link below
Manhattan challenge problem

Problem Solving - 6

For every positive integer n, the function h(n) is defined to be the product of all of the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

A. between 2 and 10

B. between 10 and 20

C. between 20 and 30

D. between 30 and 40

E. greater than 40

Answer - E

h(n)=2*4*6....100 =2(1*2*3*........50)

h(n) + 1 = 2(1*2*...50) + 1

Prime No ={2,3,........}

h(n) contains multiples of all the primes uptil 50 .... Hence E

OR

Let X = 2*4*6*.....*100

then X contains multiples of all the primes uptil 50
eg.
17 * 2 = 34 is in X.

37 * 2 = 74 is in X

Hence all the primes between 2 - 50 will be a factor of X => that none of them will be a factor of (X+1), hence the smallest prime should be at least greater than 50

Hence E.

DS Question - 11

a, b, c all are positive, is a/b > (a+c)/(b+c) ?

1. a > c

2. a > b

Answer - B

statement (1). - insufficient as we do not know the relation between a and b

statement (2). - a > b gives answer irrespective of whether ( c <> b) that, a / b > (a + c) / (b + c)

a/b > [a+c]/[b+c]

= a(b+c) > b(a+c)

= ab+ac > ab+bc

= ac > bc

=a > b

DS Question - 10

When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15 ?

(1) n- 2 is divisible by 5.

(2) t is divisible by 3.

Answer - C

Statement (1) alone - no sufficient information about t
Statement (2)alone - no sufficient information about n

From the ques --- n = a*3 + 2

t = b*5 + 3

Using Statement (1) --- n = p*5 + 2

Thus n = 15*k + 2

Using Statement (2) --- t = q*3 + 3

Thus t = 15*j + 3

nt = 15*k(15*j + 3) + 30*j + 6

=> always leaves a remainder of 6 when divided by 15.

Problem Solving - 5

If |x| > 3, which of the following must be true?

A). x>3

B). x<3

C). x=3

D). x is not equal to 3

E). x<-3

Answer - D

A,B,E - Ruled out - a modulus sign in the question=>cannot have a single signed (can always find a solution with the opposite sign )

When x is positive ----- x >3

When x is negative ---- x <-3

Thus x can never be equal to 3.

DS Question - 9

Find an integer S.

(1) S=3

(2) S^3=3^S

Answer - B

Statement (1) - insufficient ---- S=3 or –3
Statement (2) - sufficient ---- S = 3

X^Y = Y^X only if X and Y are positive integers of equal value
=> X and Y have to be positive, otherwise one side will give a negative value when X or Y is odd while other side will give a decimal value

Given that Y is equal to 3, X has to be equal to 3.

Problem Solving - 4

What is the sum of all three digit positive integers, the sum of whose digits is a multiple of 3?

a) 900

b) 165150

c) 165000

d) 329199

e) 328900

Answer - B

Sum of n numbers = n/2(2a+(n-1)d)

We want to add the numbers from 102 to 999 that are divisible by 3.

We divide by 3 to get 34 to 333, or 300 numbers in total.

Here n = 300, a = 34, d= 1

Sum = 1/2*300*(333+34) = 150 * 367 = 55,050

Multiplying by 3 again gives 165,150 .

TIP

To add the first n integers from 1 to n
add together 1 + n = n+1
then 2 + (n-1) = n + 1 until you get to n/2 (depending on whether n is odd or even)
Ultimately we have n/2 pairs that all add up to (n+1)
So the sum is 1/2 * n * (n+1)
If you don't start at 1, or have a gap of more than 1 --- slightly adjust the formula.

DS Question - 8

If X and Y are nonzero integers, what is the remainder when X is divided by Y?

(1) When X is divided by 2y, the remainder is 4

(2) When X+Y is divided by y, the remainder is 4

Answer - B

Statement (1) ---- x = (2y)M +4

if y > 4 the remainder of x/y will be 4

if y= 1 or 2 or 4 there will be no remainder,

if y=3, remainder will be 1,

so it depends on value of y...

so insufficient

Statement (2) -----(x+y)/y=M+4

It is clear that y/y=1 therefore (x/y) has to be 4 - Thus

sufficient