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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, 17:12

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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.

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

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 \(1000r+100(s+2)+10t+u\) --> \(n-m=(1000r+100(s+2)+10t+u)-1000r+100s+10t+u=200\).

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

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

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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

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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