For any four digit number

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For any four digit number

by BTGmoderatorDC » Sun Sep 10, 2017 2:07 pm
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

Hi Experts, do any idea how to solve this please? I am quite struggling and badly need help.

OA B

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by Jay@ManhattanReview » Mon Sep 11, 2017 6:33 am
lheiannie07 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

Hi Experts, do any idea how to solve this please? I am quite struggling and badly need help.

OA B
We know that *abcd*= (3^a)(5^b)(7^c)(11^d)

This means that the exponent of 3 is the thousands digit; the exponent of 5 is the hundreds digit; the exponent of 7 is the tens digit; and the exponent of 11 is the unit digit.

From *m* = (3^r)(5^s)(7^t)(11^u), we get that *m* = *rstu*

We are given that *n* = (25)(*m*).

Thus, *n* = (25) x (3^r)(5^s)(7^t)(11^u) = (5^2)*(3^r)(5^s)(7^t)(11^u) = (3^r)(5^(s+2))(7^t)(11^u)

=> *n* = *r(s+2)tu*

*m* = *rstu* = 1000r + 100s + 10t + u ---(1)
*n* = *r(s+2)tu* = 1000r + 100(s+2) + 10t + u ---(2)

Thus, from (2) - (1), we get

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

The correct answer: B

Hope this helps!

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