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Digit placement and exponents question (700-800 level)

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Digit placement and exponents question (700-800 level)

by kylecrogers » Sat Dec 07, 2013 4:25 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

The answer makes sense to me, but I don't understand the basis for one key assumption. Please help by working it through, and I will explain the assumption when we get there.

Thanks!
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by kylecrogers » Sat Dec 07, 2013 4:30 pm
Nevermind... I just got it.
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by Uva@90 » Sat Dec 07, 2013 6:24 pm
kylecrogers 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

The answer makes sense to me, but I don't understand the basis for one key assumption. Please help by working it through, and I will explain the assumption when we get there.

Thanks!
The four digit number m = rstu
the four digit number n = 25(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) => r(s+2)tu
since s is 100th digit.
their difference is 200

Answer is B ?
is it right ?

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Uva.
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by GMATGuruNY » Sun Dec 08, 2013 12:26 pm
kylecrogers 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
m:
Let m = 1000.
Then *m* = *1000* = 3¹5�7�11� = 3.

n:
*n* = (25)(*m*) = 25*3 = 75.
Thus:
n is a 4-digit integer abcd such that *abcd* = 75.

Since *abcd* = (3^a)(5^b)(7^c)(11^d), we get:
(3^a)(5^b)(7^c)(11^d) = 75
(3^a)(5^b)(7^c)(11^d) = 3¹5²7�11�
abcd = 1200.

Result:
n is 4-digit integer 1200.

Thus:
n - m = 1200-1000 = 200.

The correct answer is B.
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by aalradadi » Sun Dec 08, 2013 3:19 pm
GMATGuruNY wrote:
kylecrogers 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
m:
Let m = 1000.
Then *m* = *1000* = 3¹5�7�11� = 3.

n:
*n* = (25)(*m*) = 25*3 = 75.
Thus:
n is a 4-digit integer abcd such that *abcd* = 75.

Since *abcd* = (3^a)(5^b)(7^c)(11^d), we get:
(3^a)(5^b)(7^c)(11^d) = 75
(3^a)(5^b)(7^c)(11^d) = 3¹5²7�11�
abcd = 1200.

Result:
n is 4-digit integer 1200.

Thus:
n - m = 1200-1000 = 200.

The correct answer is B.
Hey Mitch,

If we choose any number for M , will we get the same answer?
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by GMATGuruNY » Sun Dec 08, 2013 3:54 pm
aalradadi wrote:Hey Mitch,

If we choose any number for M , will we get the same answer?
Yes, we can plug in any 4-digit integer for m and solve for n.
To illustrate:

m:
Let m = 1100.
Then *m* = *1100* = 3¹5¹7�11� = 15.

n:
*n* = (25)(*m*) = 25*15 = 375.
Thus:
n is a 4-digit integer abcd such that *abcd* = 375.

Since *abcd* = (3^a)(5^b)(7^c)(11^d), we get:
(3^a)(5^b)(7^c)(11^d) = 375
(3^a)(5^b)(7^c)(11^d) = 3¹5³7�11�
abcd = 1300.

Result:
n is 4-digit integer 1300.

Thus:
n - m = 1300-1100 = 200.
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Alternative method

by Mathsbuddy » Sun Dec 08, 2013 5:57 pm
Here's an alternative approach:

n/m = 25
(3^a * 5^b * 7^c * 11^d)/(3^r * 5^s * 7^t * 11^u) = 25
3^(a-r) * 5^(b-s) * 7^(c-t) * 11^(d-u) = 5^2
So a-r=0, b-s = 2, c-t=0, d-u=0

Also *n* - *m* = DIGITS [a-r, b-s, c-t, d-u] = [0200]

Answer B) 200
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