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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