For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). What is the value of (n - m) if m and n are four-digit numbers for which *m* = (3r)(5s)(7t)(11u) and *n* = (25)(*m*)?
A)2000
B)200
C)25
D)20
E)2
Ans: b
Source: MGMAT Practice test
GMAT Forum
4 postsPage 1 of 1
- Anand
- Anand
Functions
Yes, I am sure that question in neither incorrect nor incomplete.
Can MGMAT instructor help to understand the solution to the problem.
Can MGMAT instructor help to understand the solution to the problem.
- esledge
- Forum Guests
- Posts: 1181
- Joined: Tue Mar 01, 2005 6:33 am
- Location: St. Louis, MO
There is a critical bit of info: the variables are exponents. I'll reproduce the question here, with ^ indicating exponents.
For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What is the value of (n - m) if m and n are four-digit numbers for which *m* = (3^r)(5^s)(7^t)(11^u) and *n* = (25)(*m*)?
A 2000
B 200
C 25
D 20
E 2
This is a neat one: it combines functions, digits, and prime factors into one problem.
The asterisk function tells you what to do with the digits in a four digit number:
The thousands place indicates how many 3's to multiply together.
The hundreds place indicates how many 5's to multiply together.
The tens place indicates how many 7's to multiply together.
The units place indicates how many 11's to multiply together.
End result: some big number that may have a bunch of 3's, 5's, 7's and 11's in it's prime factorization....and NO other primes.
It is VERY important that 3, 5, 7, and 11 are all unique primes (no shared factors). Why? If the result of the asterisk function on some four digit number is 99, which factors to (3^2)(11), then we know that the function called for exactly two 3's, no 5's, no 7's and one 11. This implies that the four digit number plugged into the function was 2001.
Similarly, we can work back from the statement that *m* = (3^r)(5^s)(7^t)(11^u) to conclude that the four digits of m are r/s/t/u {I just mean for the slashes to separate digits, not anything weird.}
Since *n* = (25)(*m*) = (5^2)(*m*), we can conclude that n has two more 5's in its factorization than *m* does. The hundreds digit of n indicates the number of 5's to multiply. Thus, n's hundreds digit is two greater than m's hundreds digit. All the other digits of m and n are the same. We conclude that the four digits of n are r/(s+2)/t/u {again, slashes just separate the digits}
Take the difference n-m:
n = r / (s+2) / t / u
m = r / (s+0) / t / u
n-m= 0 / 2 / 0 / 0 = two hundred
Of course, if you recognized this concept from previous practice (or see it again in the future), you can take a shortcut. Since *n* is given in terms of *m*, and they only differ by that factor of 25, the only difference between n and m is in the hundreds place. Eliminate all answers but (B). Additionally, it makes sense for 2 to be in the hundreds place, as that is the number of 5's in 25.
For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What is the value of (n - m) if m and n are four-digit numbers for which *m* = (3^r)(5^s)(7^t)(11^u) and *n* = (25)(*m*)?
A 2000
B 200
C 25
D 20
E 2
This is a neat one: it combines functions, digits, and prime factors into one problem.
The asterisk function tells you what to do with the digits in a four digit number:
The thousands place indicates how many 3's to multiply together.
The hundreds place indicates how many 5's to multiply together.
The tens place indicates how many 7's to multiply together.
The units place indicates how many 11's to multiply together.
End result: some big number that may have a bunch of 3's, 5's, 7's and 11's in it's prime factorization....and NO other primes.
It is VERY important that 3, 5, 7, and 11 are all unique primes (no shared factors). Why? If the result of the asterisk function on some four digit number is 99, which factors to (3^2)(11), then we know that the function called for exactly two 3's, no 5's, no 7's and one 11. This implies that the four digit number plugged into the function was 2001.
Similarly, we can work back from the statement that *m* = (3^r)(5^s)(7^t)(11^u) to conclude that the four digits of m are r/s/t/u {I just mean for the slashes to separate digits, not anything weird.}
Since *n* = (25)(*m*) = (5^2)(*m*), we can conclude that n has two more 5's in its factorization than *m* does. The hundreds digit of n indicates the number of 5's to multiply. Thus, n's hundreds digit is two greater than m's hundreds digit. All the other digits of m and n are the same. We conclude that the four digits of n are r/(s+2)/t/u {again, slashes just separate the digits}
Take the difference n-m:
n = r / (s+2) / t / u
m = r / (s+0) / t / u
n-m= 0 / 2 / 0 / 0 = two hundred
Of course, if you recognized this concept from previous practice (or see it again in the future), you can take a shortcut. Since *n* is given in terms of *m*, and they only differ by that factor of 25, the only difference between n and m is in the hundreds place. Eliminate all answers but (B). Additionally, it makes sense for 2 to be in the hundreds place, as that is the number of 5's in 25.
Emily Sledge
Instructor
ManhattanGMAT
Instructor
ManhattanGMAT
4 posts Page 1 of 1