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

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19 Mar 2007, 21:49

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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*)?

What I mean by absorbing is this : when you absorb 25 in 5^5 the result is 5^7. So power is gone up by 2.

Now if you read the line in the earlier post again, ( modified it a bit )

Quote:

So, for *n* ( which is 25 times *m*) , if you absorb the 25 as 5^2 within *m*, then the value for s in (5s) in eq [2] will go up by 2

You will see here value of s is going up by 2. S is digit at 100th place in value of m [*m* = (3r)(5s)(7t)(11u)]. So when s goes up by 2, the value is actually gone up by 2*100. (similarly when r goes up by 2 value goes up by 2000 etc..)

For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). What is the value of (n â€“ m) if m and n are four-digit numbers for which *m* = (3r)(5s)(7t)(11u) and *n* = (25)(*m*)?

a)2000 b)200 c)25 d)20 e)2

(3a) means 3 to the power a... (5b) means 5 to the power b... and so on...

The answer is B.

"The fundamental theorem of arithmetic states that every positive integer larger than 1 can be written as a product of one or more primes in a unique way, i.e. unique except for the order. The same prime may occur multiple times. "

Lets set f(m) = *m* = *(rstu)*=(3r)(5s)(7t)(11u)
and f(n) = *n* = *(xyzq)*= (3x)(5y)(7z)(11q)

we see that 3,5,7,11 - prime numbers.

but from other hand f(n) = (25)(*m*) = (25)(3r)(5s)(7t)(11u) = (5^2)(3r)(5s)(7t)(11u) = (3r)(5^s+2)(7t)(11u)

thus, x=r, y=s+2, z=t and u=q. the difference is only y=s+2 - the hundred's digit.
it meas that n - m = |r| |s+2| |t| |u| - |r| |s| |t| |u| = 200.

There was a typo in the stem. Original question should read:

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

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.

In case of any question please continue discussion here: abcd-126522.html