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]

24 Jan 2012, 17:12

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Question Stats:

53% (02:57) correct

46% (01:31) wrong

based on 26 sessions

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, 04:28, edited 2 times in total.

OA added

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

25 Jan 2012, 04:28

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enigma123 wrote:

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: *abcd* [#permalink]

24 Jan 2012, 17:27

enigma123 wrote:

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.
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Director

Status: Preparing for the 4th time -:(

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Kudos [?]: 71 [0], given: 212

Re: *abcd* [#permalink]

24 Jan 2012, 17:31

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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Re: *abcd* [#permalink]

24 Jan 2012, 17:37

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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Re: *abcd* [#permalink]

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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Challenging Digits Q [#permalink]

10 Jan 2013, 09:23

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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Re: Challenging Digits Q [#permalink]

10 Jan 2013, 16:19

sambam 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

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Re: Challenging Digits Q [#permalink] 10 Jan 2013, 16:19

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

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