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

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enigma123
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For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   Updated on: 01 Feb 2020, 02:18
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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
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B

Originally posted by enigma123 on 24 Jan 2012, 17:12.
Last edited by Bunuel on 01 Feb 2020, 02:18, edited 3 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). What [#permalink] New post   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 \(m\): \(n-m=200\).

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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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   25 Jan 2012, 04:28
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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). What [#permalink] New post   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: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   24 Jan 2012, 17:37
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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: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   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). What [#permalink] New post   14 Dec 2015, 08:33
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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
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   02 Sep 2017, 01:08
Bunuel wrote:
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 \(m\): \(n-m=200\).

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.



n-m = 24m
why 200 is not divisible by 24. What am i missing
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   02 Sep 2017, 03:16
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abrakadabra21 wrote:
Bunuel wrote:
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 \(m\): \(n-m=200\).

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.



n-m = 24m
why 200 is not divisible by 24. What am i missing


How did you get that n - m = 24? We are given that *n* = (25)(*m*), not that n = 25m. Notice that both n and m are functions (*n* and *m*, not n and m).
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   05 Sep 2018, 08:54
This is a good question as it is tricky with many variables
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   18 Mar 2019, 15:23
+1 for plugging in... once we get that the 5s are the hundreds place, just make every variable 0 so each place value is now 1.
So, we get n-m = 1(1+2)11 - 1111 = 200
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   23 Mar 2019, 09:22
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.



n=25m
n-m=25m=m=24m= 8 * 3m
ie., m is a multiple of 8
2000 and 200 are multiples of 8
Now,
2000=250/3=(25*10)/3=(5^3)*2*(3^-1)
But m is not a factor of 2 as defined in the question.
200=25/3=(5^2)*(3^-1)
Therefore,
answer is B
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink] New post   25 Oct 2023, 22:52
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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


Responding to a pm: The tricky thing about this question is to understand that we are using an operator on a 4 digit number abcd.

For a 4 digit number abcd, \(*abcd*= (3^a)(5^b)(7^c)(11^d)\) similar to \(f(abcd) = (3^a)(5^b)(7^c)(11^d)\)

Then given \(*m* = (3^r)(5^s)(7^t)(11^u) = f(rstu)\) it means m = rstu

\(*n* = (25)(*m*) = 25 * (3^r)(5^s)(7^t)(11^u) = (3^r)(5^{s+2})(7^t)(11^u) \) so n = r(s+2)tu i.e. the hundreds digit of n is 2 more than that of m.

Now assume m = 1111, then n = 1311 and the difference between them is simply 200.

Answer (B)

Check this video on Functions: https://youtu.be/ik5XijHCOAI
and this post: https://anaprep.com/algebra-functions/
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Re: For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What [#permalink]
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