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For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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19 Mar 2007, 21:49
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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: foranyfourdigitnumberabcdabcd3a5b7c11d126522.html
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Last edited by Bunuel on 13 Feb 2012, 04:29, edited 2 times in total.
Edited the question and added the OA. Topic is locked.



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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
NM = 200



Senior Manager
Joined: 29 Jan 2007
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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)



Director
Joined: 13 Dec 2006
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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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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.



Manager
Joined: 04 Jan 2006
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Kyatin,
According to your description, 200 is *n*  *m*
Question is to find nm



Senior Manager
Joined: 29 Jan 2007
Posts: 441
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nope...I disagree. My answer is indeed nm,
Please give it a shot...I will explain when you give up
fight!



Manager
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Re: PS: Asterisks away [#permalink]
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20 Mar 2007, 16:18
faifai0714 wrote: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). What is the value of (n â€“ m) if m and n are fourdigit 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.



Director
Joined: 13 Dec 2006
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Yes Kyatin it does make sense.... thanks for the explanation buddy
regards,
Amardeep



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Joined: 16 Dec 2011
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Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). [#permalink]
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13 Feb 2012, 03:13
need explanation for this thanks



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Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). [#permalink]
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13 Feb 2012, 04:24
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 \(m\): \(nm=200\). Answer: B. Or represent four digits integer \(rstu\) as \(1000r+100s+10t+u\) and four digit integer \(r(s+2)tu\) as \(1000r+100(s+2)+10t+u\) > \(nm=(1000r+100(s+2)+10t+u)1000r+100s+10t+u=200\). Answer: B. In case of any question please continue discussion here: abcd126522.htmlHope it helps.
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Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d).
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13 Feb 2012, 04:24







