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:
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?
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`
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`.
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`.
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`
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.
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`
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)) ...`
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`
