GRE & GMAT Mathy Tutor

Tuesday, July 24, 2012

GMAT Quantitative Scoring, isn't 60 the max?

Let's start with the official scoring table from the horse's mouth:

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:

Try this word problem

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?

how many factors?

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:

$8*3*2*2 = 96$

Sunday, April 1, 2012

Sum of two 2-digit numbers?

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:

$M = 10*a + b$

$N = 10*b + a$
(adding these two equations)

$M + N = 11*a + 11*b = 11*(a + b)$
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

$N$ .

Thursday, March 1, 2012

How Many Non-Zero Digits?

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

$64$ .

Wednesday, February 1, 2012

What number was removed?

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:

Sunday, January 1, 2012

GRE Factoring Problem

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

$2$ s in

$n$ , which is

$2^11$ which is

$2048$ , (actually there will be far more

$2$ s needed ).
if we focused on

$3$ s, I see 6 multiples of

$3$ , meaning we'd at least need 7

$3$ s 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.

$n$ =

$46$ .