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

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Manager
Joined: 02 Sep 2006
Posts: 221
For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d  [#permalink]

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Updated on: 13 Feb 2012, 04:29
1
8
00:00

Difficulty:

75% (hard)

Question Stats:

63% (01:06) correct 37% (01:40) wrong based on 150 sessions

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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: for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html

This topic is locked. MODERATOR.

Originally posted by faifai0714 on 19 Mar 2007, 21:49.
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.
VP
Joined: 22 Oct 2006
Posts: 1334
Schools: Chicago Booth '11

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20 Mar 2007, 06:55
1
1
1
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

N-M = 200
Senior Manager
Joined: 29 Jan 2007
Posts: 363
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20 Mar 2007, 08:55
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)
Senior Manager
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Posts: 396
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20 Mar 2007, 09:23
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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Joined: 29 Jan 2007
Posts: 363
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20 Mar 2007, 13:41
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.
Intern
Joined: 04 Jan 2006
Posts: 38

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20 Mar 2007, 14:00
Kyatin,
According to your description, 200 is *n* - *m*
Question is to find n-m
Senior Manager
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20 Mar 2007, 14:31
nope...I disagree. My answer is indeed n-m,

Please give it a shot...I will explain when you give up

fight!
Manager
Joined: 20 Jun 2005
Posts: 108
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 four-digit 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.
Senior Manager
Joined: 13 Dec 2006
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22 Mar 2007, 11:15
Yes Kyatin it does make sense.... thanks for the explanation buddy

regards,

Amardeep
Intern
Joined: 16 Dec 2011
Posts: 37
GMAT Date: 04-23-2012
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
Math Expert
Joined: 02 Sep 2009
Posts: 58454
Re: For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d).  [#permalink]

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13 Feb 2012, 04:24
4
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 $$m$$: $$n-m=200$$.

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

In case of any question please continue discussion here: abcd-126522.html

Hope it helps.
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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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30 Oct 2018, 04:39
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Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d   [#permalink] 30 Oct 2018, 04:39
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