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For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d

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For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink] New post 19 Mar 2007, 20:49
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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

Open discussion of this question is here: for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html

This topic is locked. MODERATOR.
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Feb 2012, 03:29, edited 2 times in total.
Edited the question and added the OA. Topic is locked.
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 [#permalink] New post 20 Mar 2007, 05:55
I get B


We need to substitute values to get these confusing mess simpler

We need to get as many 0's and 1's in the exponents to make it simpler.

I chose M to be 1000 since it is 4 digits, and it contains the most 0's


so since M = 1000

*M* = (3^1) (5^0) ( 7^0 ) (11 ^ 0)

this leaves as as *M* = 3

so now we know *N* = 25 (*M*)

*N* = 25(3)

*N* = 75

so to find N we need to work backwards

3^R 5^S 7^T 11^ U = 75

what values of R S T U would yield 75? 7 and 11 are out since they are not factors so T and U get 0

3 and 5 to make 75 would be 3^1 * 5 ^2 = 75

so the first 2 digits would be 1 and 2 , last 2 would be 0 and 0 and we get

N = 1200

M = 1000

N-M = 200
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 [#permalink] New post 20 Mar 2007, 07:55
I got 200 (B) as well.
But did it with a different approach

We are given *n* = 25 x *m* -eq [1]
and also *m* = (3r)(5s)(7t)(11u) -eq [2]

So, for *n*, if you absorb the 25 as 5^2 within *m*, then the value for s in (5s) in eq [2] will go up by 2

So if *m* = 1000 then *n* = 1200

So answer is 200 (B)
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 [#permalink] New post 20 Mar 2007, 08:23
Kyatin, apparently your answer is smart,
can you please explain following line to me

So, for *n*, if you absorb the 25 as 5^2 within *m*, then the value for s in (5s) in eq [2] will go up by 2

regards,

Amardeep
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 [#permalink] New post 20 Mar 2007, 12:41
Amardeep

What I mean by absorbing is this : when you absorb 25 in 5^5 the result is 5^7. So power is gone up by 2.

Now if you read the line in the earlier post again, ( modified it a bit )

Quote:
So, for *n* ( which is 25 times *m*) , if you absorb the 25 as 5^2 within *m*, then the value for s in (5s) in eq [2] will go up by 2


You will see here value of s is going up by 2. S is digit at 100th place in value of m [*m* = (3r)(5s)(7t)(11u)]. So when s goes up by 2, the value is actually gone up by 2*100. (similarly when r goes up by 2 value goes up by 2000 etc..)

Let me know if it makes sense.
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 [#permalink] New post 20 Mar 2007, 13:00
Kyatin,
According to your description, 200 is *n* - *m*
Question is to find n-m
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 [#permalink] New post 20 Mar 2007, 13:31
nope...I disagree. My answer is indeed n-m,

Please give it a shot...I will explain when you give up :lol:

fight!
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Re: PS: Asterisks away [#permalink] New post 20 Mar 2007, 15:18
faifai0714 wrote:
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

(3a) means 3 to the power a...
(5b) means 5 to the power b...
and so on...


The answer is B.

"The fundamental theorem of arithmetic states that every positive integer larger than 1 can be written as a product of one or more primes in a unique way, i.e. unique except for the order. The same prime may occur multiple times. "

Lets set f(m) = *m* = *(rstu)*=(3r)(5s)(7t)(11u)
and f(n) = *n* = *(xyzq)*= (3x)(5y)(7z)(11q)

we see that 3,5,7,11 - prime numbers.

but from other hand f(n) = (25)(*m*) = (25)(3r)(5s)(7t)(11u) = (5^2)(3r)(5s)(7t)(11u) = (3r)(5^s+2)(7t)(11u)

thus, x=r, y=s+2, z=t and u=q. the difference is only y=s+2 - the hundred's digit.
it meas that n - m = |r| |s+2| |t| |u| - |r| |s| |t| |u| = 200.
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 [#permalink] New post 22 Mar 2007, 10:15
Yes Kyatin it does make sense.... thanks for the explanation buddy :-D

regards,

Amardeep
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Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). [#permalink] New post 13 Feb 2012, 02:13
need explanation for this thanks
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Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). [#permalink] New post 13 Feb 2012, 03:24
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pbull78 wrote:
need explanation for this thanks


There was a typo in the stem. Original question should read:

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

Given for four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d;

From above as *m*=3^r*5^s*7^t*11^u then four digits of m are rstu;

Next, *n*=25*\{*m*\}=5^2*(3^r*5^s*7^t*11^u)=3^r*5^{(s+2)}*7^t*11^u, hence four digits of n are r(s+2)tu, note that s+2 is hundreds digit of n;

You can notice that n has 2 more hundreds digits and other digits are the same, so n is 2 hundreds more than m: n-m=200.

Answer: B.

Or represent four digits integer rstu as 1000r+100s+10t+u and four digit integer r(s+2)tu as 1000r+100(s+2)+10t+u --> n-m=(1000r+100(s+2)+10t+u)-1000r+100s+10t+u=200.

Answer: B.

In case of any question please continue discussion here: abcd-126522.html

Hope it helps.
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Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d).   [#permalink] 13 Feb 2012, 03:24
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