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For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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24 Jan 2012, 17:12
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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 Guys  any idea how to solve this please? I am struggling and OA is not given either.
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Last edited by Bunuel on 13 Feb 2012, 04:28, edited 2 times in total.
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Re: *abcd* [#permalink]
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24 Jan 2012, 17:27
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enigma123 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
Guys  any idea how to solve this please? I am struggling and OA is not given either. 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.
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Re: *abcd* [#permalink]
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24 Jan 2012, 17:31
Sorry buddy  apologies if I am missing something. But how did you get four digits of m as rstu;
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Re: *abcd* [#permalink]
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24 Jan 2012, 17:37



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Re: *abcd* [#permalink]
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24 Jan 2012, 17:42
Ahhh  so four digits are r, s, t and u. And not r minus s minus t minus u. That's where I got confused.
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Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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25 Jan 2012, 04:28
enigma123 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
Guys  any idea how to solve this please? I am struggling and OA is not given either. Also, if you find to difficult to grasp a question with many variables, try throwing in some values. It helps you handle the question. abcd is a four digit number where a, b, c and d are the 4 digits. *abcd*= (3^a)(5^b)(7^c)(11^d). The '**' act as an operator. Given: *m* = (3^r)(5^s)(7^t)(11^u) So m = rstu where r, s, t, and u are the 4 digits of m. Say, r = 1 and s = 0, t = 0 and u = 0 m = 1000 Then *m* = 3 Now, *n* = (25)(*m*) = 25(3) = (3^1)(5^2)(7^0)(11^0) n = 1200 n  m = 1200  1000 = 200
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Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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14 Dec 2015, 08:33
Hi Bunuel Can we arrive at the solution by the following approach ? Given: *m* = 3^r*5^s*7^t*11^u *n* = 25 (*m*) To Solve: n  m Sol: Substituting for n , n  m = 25 *m*  *m* = *m* (251) = *m* (24) we know that, 24 = 3*2^3 and *m* = 3^r*5^s*7^t*11^u , does not have 2 value which implies the answer should have 2^3 as a factor. 1. 2000 = 5^3*2^4  ( Only 2^3 is possible. as 24 has only 2^3 and *m* is not a factor of 2) 2. 200 = 5^2*2^3  Correct 3. 25 = 5^2 4. 20 = 5 *2^2 5. 2 = 2



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Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
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13 Mar 2017, 12:06
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Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d
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