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For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
24 Jan 2012, 16:12

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Difficulty:

35% (medium)

Question Stats:

62% (02:41) correct
38% (01:54) wrong based on 102 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.

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.

Re: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
25 Jan 2012, 03:28

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This post received KUDOS

Expert's post

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

Challenging Digits Q [#permalink]
10 Jan 2013, 08:23

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*)?

Re: Challenging Digits Q [#permalink]
10 Jan 2013, 15:19

Expert's post

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: For any four digit number, abcd, *abcd*=3^a*5^b*7^c*11^d [#permalink]
05 Mar 2014, 09:12

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