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700-800 MGMAT

by pkw209 » Mon May 10, 2010 1:51 pm
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
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by aturpening » Mon May 10, 2010 6:46 pm
The answer is b.

When you multiply *m* by 25, it's the same thing as multiplying it by 5^2. 5^2 would fall into the second part of the term, 5^b and corresponds with the hundreds digit of the number m and n. If this is the only difference between the two numbers, then m-n would end up being 200. b.
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by liferocks » Mon May 10, 2010 6:47 pm
m=1000r+100s+10t+u

*n*=5^2*(3^r)(5^s)(7^t)(11^u)=(3^r)(5^(s+2))(7^t)(11^u)

hence n=1000r+100(s+2)+10t+u

n-m=200

ans option B
"If you don't know where you are going, any road will get you there."
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by pkw209 » Mon May 10, 2010 7:14 pm
How did you know that 3^r for example corresponds to the thousands digit? also, would it be possible to elaborate on the breakdown a little more?

Could you also solve by plugging in numbers?
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by pkw209 » Mon May 10, 2010 7:15 pm
also, how did you jump from 5^s+2 to 100(s+2)?
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by liferocks » Mon May 10, 2010 7:25 pm
pkw209 wrote:also, how did you jump from 5^s+2 to 100(s+2)?
In the question it is given that
*abcd*= (3^a)(5^b)(7^c)(11^d)

now for abcd ,a is the digit at thousandth place and so on...
if we map the same logic for *m* = (3^r)(5^s)(7^t)(11^u) i.e let m=abcd then
*m*=(3^a)(5^b)(7^c)(11^d)=(3^r)(5^s)(7^t)(11^u)..shows a=r or r is the digit at thousands place

as per number picking...even if we pickup numbers finally we have to get each digits of the number
ex let m=1000..then *m*=3
*n*=75=3^1*5^2..so n=1200
n-m=200
"If you don't know where you are going, any road will get you there."
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by pkw209 » Mon May 10, 2010 7:48 pm
In the question it is given that
*abcd*= (3^a)(5^b)(7^c)(11^d)

now for abcd ,a is the digit at thousandth place and so on...
if we map the same logic for *m* = (3^r)(5^s)(7^t)(11^u) i.e let m=abcd then
*m*=(3^a)(5^b)(7^c)(11^d)=(3^r)(5^s)(7^t)(11^u)..shows a=r or r is the digit at thousands place

as per number picking...even if we pickup numbers finally we have to get each digits of the number
ex let m=1000..then *m*=3
*n*=75=3^1*5^2..so n=1200
n-m=200
That's the explanation I was looking for. Extremely helpful.

Thanks!
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by sanju09 » Tue May 11, 2010 2:41 am
pkw209 wrote: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
In here, m is the four-digit number rstu, and n is another four-digit number such that

*n* = (3^r){5^(s + 2)}(7^t)(11^u), or n's four digits are r(s + 2)tu, s is such that s + 2 is still a digit, now imagine the subtraction of rstu from r(s + 2)tu, all places find a 0 EXCEPT the hundreds' place, which will be 2, and hence the difference n - m will be [spoiler]200.

B[/spoiler]
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by kstv » Tue May 11, 2010 3:27 am
m can be a 4 digit no = 1111 ,

*m* = 3*5*7*11 cos 3^1 = 3

*n* = 25(*m*) = 5²(3*5*7*11)

3*5^(2+1) *7*11 = 3*5^3*7*11

so n = 1311

n-m = 1311 - 1111 = 200

Picking no is easier.
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by pkw209 » Tue May 11, 2010 7:40 am
fantastic. thanks for the responses! :)
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