# Manhattan/Williamsburg GRE & GMAT Math Tutor

2012-07-24

The Quantitative Score range is 0 - 60, and it says that scores over 50 are extremely rare.  I consequently was shooting for a 60 when I took the test on June 6th 2012.  I was dismayed to receive this score report:

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2012-07-03

Laura is 3 times as old as Maria was when Laura was as old as Maria is now. In 2 years Laura will be twice as old as Maria was 2 years ago. How old are they now?

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2012-05-01

How many unique factors does 40320 have?

[Spoiler Below]

Video:

Text:

The prime factorization of 40320 yields:
40320 = 2^7*3^2*5^1*7^1

so any factor of of 40320 can be written in this form:

2^a*3^b*5^c*7^d
where a,b,c, and d are all non-negative integers and:
a<=7
b<=2
c<=1
d<=1

so the number of options for a is 8 (including 0), and for b: 3, c: 2, and d: 2.
so to count all the permutations of factors, we take the product:
8*3*2*2 = 96

2012-04-01

If the two-digit integers M and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of N and M?

A) 181
B) 165
C) 121
D) 99
E) 44

[Spoiler Below]

Without loss of generality, le's say M has a as the tens digit, and b as the units digit.
Consequently, N would have b as the tens, and a as the units digit.

remembering how the decimal system works:
M = 10*a + b
N = 10*b + a
M + N = 11*a + 11*b = 11*(a + b)
since a and b are both integers, a + b must also be an integer, so M + N = 11* an_integer
so M + N is a multiple of 11.

A) 181 is prime
B) 165 = 11 * 5 * 3
C) 121 = 11 * 11
D) 99 = 11 * 3^2
E) 44 = 11 * 4

(A) is the only choice that isn't a multiple of 11, and therefore cannot be the the sum of M and N.

2012-03-01

How many nonzero digits does 1/(2^13 * 5^19) have when written as a decimal?

[Spoiler Below]

1/(2^13*5^19) = 1/(2^13*5^13*5^6) = 1/((2*5)^13*5^6) = 1/(10^13*5^6) = 10^-13 * 1/(5^6)
note I can multiple by 1 whenever I want without changing the value, and 1 = (2^6/2^6), continuing:
 = 10^-13 * (2^6/2^6)*1/5^6
 = 10^-13 * 2^6/(2^6*5^6)
 = 10^-13 * 2^6/(2*5)^6
 = 10^-13*2^6/10^6 = 10^-13 * 10^-6 * 2^6 = 2^6 * 10^-19
 = (2^3)^2 * 10^-19 = 8^2 *10^-19 = 64 * 10^-19

the 10^-19 is only going to change the place of the decimal, not the number of non-zero digits, so the answer is simply 2, the number of digits in 64.

2012-02-01

A Basic GRE/GMAT Word problem:

The average of 8 numbers is 20.  If after one number is removed, the average of the remaining 7 numbers is 16.  What number was removed?

[Spoiler Below]

Remembering that the mean is the sum divided by the number of numbers:
1: "Average"_8 = 20 = "Sum"_8/8
2: "Average"_7 = 16 = "Sum"_7/7
so:
3: 160 = "Sum"_8
and:
4: 112 = "Sum"_7

noting that: "Sum"_8 = "Sum"_7 + RemovedNumber,
5: "Sum"_8 = 160 = "Sum"_7 + RemovedNumber

Subtracting (4) from (5), we get: 48 = RemovedNumber

Video:

2012-01-01

What is the smallest non-prime integer that isn't a factor of  20!?

[Spoiler Below]

Since the solution to this puzzle (let's call it n) isn't a factor of 20!, it must either:
1. Have more of a specific prime in its prime factorization than 20! does.
2. Have a prime factor that isn't in the prime factorization of 20!

Let's examine what #1 yields:
since 20! = 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1,
we can see 10 even numbers there, so at the very least we would need 11 2s in n, which is 2^11 which is 2048, (actually there will be far more 2s needed ).
if we focused on 3s, I see 6 multiples of 3, meaning we'd at least need 7 3s in n, 3^7 which is 2187
I see 4 multiples of 5, so we would need 5^5 which is 3125
2 multiples of 7, which would require 7^3 which is 343
and the smallest prime that only occurs once in 20! is 11, which would mean we'd need 11^2, which is 121.

so the smallest potential n value we get from this approach is 121, let's think about #2:
let's choose the smallest prime that isn't a factor of 20!, to find it we can start with 20 and increment until we get a prime:
21=3*7
22=11*2
23 is prime

since 23 is prime, and larger than 20, it can't be in the prime factorization of 20!, if that's not clear then look at how we wrote out 20! above into a product of decreasing integers... each of those integers is less than 20, therefore none of those integers has 23 as a factor, and therefore the product of all those integers doesn't have 23 in its prime factorization.

but the problem is the find the smallest non-prime non-factor of 20!, so we need to make it not-prime.

Since 23 is not a factor of 20! any multiple of 23 will also not be a factor of 20!, the smallest multiple of 23 which is greater than 23 is 2*23=46, this is no longer prime, but definitely isn't a factor of 20!.

this is much better (smaller) than 121, and therefore must be the solution.

n = 46.

2011-12-01

If x+y=a and x-y=b, then 2*x*y=
(A) (a^2-b^2)/2
(B) (b^2-a^2)/2
(C) (a-b)/2
(D) (a*b)/2
(E) (a^2+b^2)/2

A classic problem that I see all the time on both the GMAT and GRE, take a look at one way to solve it:
[spoiler below]

we start out with two equations:
1: x+y=a
2: x-y=b
after squaring both sides of (1):
3: (x+y)^2=x^2 + 2*x*y + y^2 = a^2
after squaring both sides of (2):
4: (x-y)^2=x^2 -2*x*y+y^2=b^2
now lets subtract (4) from (3) to get:
5: (x^2 + 2*x*y + y^2) - (x^2 -2*x*y+y^2) = 4*x*y = a^2 - b^2
now if we divide both sides of (5) by 2 we get:
6: 2*x*y=(a^2-b^2)/2

2011-11-01

What's the second smallest integer with a units digit of 5, which is the square of an integer, and the cube of an integer?

[Spoiler Below]

Video:

Text:
Let's call the number we are looking for n.

we know it's the square of an integer, so:
n = "int"_a^2
and we know it's the cube of an integer, so:
n = "int"_b^3

since any integer can be prime-factorized, let's consider the prime factorization of "int"_1 and "int"_2:
"int"_a = "prime"_1^("int"_(a1)) * "prime"_2 ^("int"_(a2)) ...

"int"_b = "prime"_1^("int"_(b1)) * "prime"_2 ^("int"_(b2)) ...

n = ("int"_a)^2 = ("prime"_1^("int"_(a1)) * "prime"_2^("int"_(a2))...)^2 = "prime"_1^(2*"int"_(a1)) * "prime"_2^(2*"int"_(a2))...

and

n = "int"_b^3 = ("prime"_1^("int"_(b1)) * "prime"_2^("int"_(b2))...)^3 = "prime"_1^(3*"int"_(b1)) * "prime"_2^(3*"int"_(b2))...

so "prime"_1^(2*"int"_(a1))*"prime"_2^(2*"int"_(a2))... = "prime"_1^(3*"int"_(b1))*"prime"_2^(3*"int"_(b2))...
and so we can say:
"prime"_x^(2*"int"_(ax)) = "prime"_x^(3*"int"_(bx)) for all x.
and therefore: 2*"int"_ax = 3*"int"_bx this is the number of "prime"_x in the prime factorization of n.
since it has a factor of 2 and 3, then the number of "prime"_x in n must be a multiple of 6.

to summarize, for any prime factor of n,there are a multiple of 6 of them in n.

we also know that the units digit of n is 5, which means it has a factor of 5, so:
n = 5 * "int"_3

so n has at least one 5 in it's prime factorization, and consequently it must have at least 6.

so 5^6 is the least integer that fits the description, but we are looking for the 2nd smallest.

well 5^12 would fit the description, and it's not the smallest, but maybe it's not the 2nd smallest,

5^6 * 2^6 is worth considering, but 5^6 * 2^6 = 10^6, which is one million, which doesn't have a 5 as the units digit.  This does not fit the description in the problem.

5^6 * 3^6 fits the description and is the second smallest integer that does.
n = 5^6 * 3^6 = 11390625