For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d : GMAT Problem Solving (PS)
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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]

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24 Jan 2012, 16:12
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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.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 13 Feb 2012, 03:28, edited 2 times in total.
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24 Jan 2012, 16:27
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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 $$m$$: $$n-m=200$$.

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

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24 Jan 2012, 16:31
Sorry buddy - apologies if I am missing something. But how did you get

four digits of m as r-s-t-u;
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24 Jan 2012, 16:37
enigma123 wrote:
Sorry buddy - apologies if I am missing something. But how did you get

four digits of m as r-s-t-u;

For four digit integer $$abcd$$ some function, denoted by **, defined as $$*abcd*=3^a*5^b*7^c*11^d$$.

Now, as given that $$*m*=3^r*5^s*7^t*11^u$$ then four digits of m are $$rstu$$, the same way as above: $$*rstu*=3^r*5^s*7^t*11^u$$

Hope it's clear.
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24 Jan 2012, 16:42
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, 03:28
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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.

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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05 Mar 2014, 09:12
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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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05 Apr 2015, 21:18
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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, 07:33
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* (25-1)
= *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
For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d   [#permalink] 14 Dec 2015, 07:33
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